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1 Integer Solutions, Rational solutions of the equations x y z x y y z z x n 4 4 and x y z xy xz y z n; And Crux Mathematicorum Contest Corner problem CC4 Konstantine Zelator P.O. Box 480 Pittsburgh, PA 503 U.S.A. address: konstantine_zelator@yahoo.com Konstantine Zelator Department of Mathematics 30 Thankeray Hall University of Pittsburgh Pittsburgh, PA 560 U.S.A. address: spaceman@pitt.edu

2 . Introduction In the May 03 issue of the journal Crux Mathematicorum (vol. 38, No 5, see reference []), the following problem appeared: (Contest Corner) CC4. Show that the equation x y z x y y z z x 4 has no integer solutions.. Does it have any rational solutions? Find one, or show that the above equation has no solutions. Contest Corner problem CC4 is the motivation behind this work. In this article, we study the equation, in three variables x,y,z; x y z x y y z z x n, () where n is a fixed or given positive integer. We provide a soluition to the first questions of CC4, in Theorem : we show that if n is of the form, n=8n, where N is an odd positive integer then equation () has no integer solutions. So question above, is really a particular case of Theorem ; the case N=3. Using Theorem, we answer question of CC4. Theorem postulates that if the integer n satisfies certain divisor conditions; then rational solutions do exist. Based on Theorem, for n=4, we find the 5 rational solution ( x, y, z) (,,) Obviously, since equation () is symmetric with respect to the three variables and since all the exponents are even; one really obtains 48=(6)(8) rational solutions from the above solution: 5 5 5,,,,,,,, ; 5 5 5,,,,,,,,. With all the sign combinations being possible (see Remark for more details). In Theorem 3, we prove a necessary and sufficient criterion for equations () to have an integer solution. Based on Theorem 3, in Theorem 4 we prove that equation () has no integer solutions if n=p (a prime), n=4, or n=p q; where p and q are distinct primes. Theorem 5 gives, in essence, a method for finding the nonnegative integer solutions of equation (). In Theorem 6, we find all the integer solutions to equation (), twenty four in total; in the case an odd prime. n p, p In Theorem 7, we determine the integer and the rational solutions of equation () in the case positive integer; and with two among x, y, z being zero. n k, k a Theorem 8 lays the case n k, k ; k. and with exactly one of the three variables x, y, z being zero. All such integer solutions are determined.

3 Similarly, Theorem 9 deals with the case such rational solutions are determined. n k, k and with exactly one of x,y,z being zero. All Theorem 0 deals with the case n=. All integer and rational solutions are listed. Theorems through 3 deal with the equation (quadratic in x), 4 4 x y z xy xz y z n In Theorem, the integer solutions of this equation (equation ()), are described in detail. Theorem states that if n= or 3(mod4); then the above (quadratic in x) equation, has no integer solutions. Finally, Theorem 3 parametrically describes all the rational solutions of equation ().. Three Lemmas Lemma Over the integers, and also over the rationals; equation (), and equations () and (3) below, have the same solution set. In other words equations (), (), and (3) are equivalent both over the rationals and over the integers. x y zx y zx y zx y z n (3) x y xy z x y xy z n () Proof If we replace the term we obtain, x y by 4x y x y in equation () x y z x y y z z x x y n ( x y z ) ( xy) n; 4 ; x y z xy x y z xy n, which is equation () Further, x y z x y z n; ( x y z)( x y z)( x y z)( x y z) n; which is equation (3) Lemma Let A, B, C be real numbers. Then, the three-variable linear system x y z A x y z B, has a unique solution; that x y z C B C A B A C solution being (x, y, z) = (x, y, z) =,, 3

4 Proof A routine calculation shows that the determinant of the matrix of the coefficients is equal to 4: D det - 4, which is not zero. Therefore the given linear system has a unique solution: We - apply Cramer s Rule. We have, A A Dx det B - ( B C), D det B y ( A B), and C - C - A Dz det - B ( A C). Therefore the unique solution is given by, C D D x B C y A B D x, y=, z A C z D D D Lemma 3 Let A, B, C, D be real numbers. Consider the 3-variable linear system, x y z A x y z B x y z C x y z D (i) If the real numbers, A, B, C, D satisfy the condition, A+D=B+C; then the given system has the unique solution B C A B A C ( x, y, z),, (ii) If A+D is not equal to B+C; the above system has no solution Proof B C A B A C (i) A routine calculation shows that since A+D=B+C; the triple,, satisfies the last or fourth equation, which is x y z D. By lemma, we also know that the same triple is the unique solution of the sub-system consisting of the first three equations. Since every solution of the system of four equations; is also a solution of any sub-system of equations; it follows that above triple is the unique solution of the system of four equations 4

5 (ii) Since any solution of the system oif four equations must be solution to the sub-system of the first three equations, it follows by Lemma, that the given system (of four equation) can have at most one solution, depending on whether the fourth equation is satisfied or not. Thus, if A D B C; the fourth equation is not satisfied and so the system has no solution 3. Theorems and, and a solution to Contest problem CC4 Theorem Suppose that n is a positive integer which is exactly divisible by 8. In other words, n=8n, where N is an odd positive integer. Then the equation, 4 4 x y z x y y z z x n () has no integer solutions Proof First observe that every integer solution to (), must satisfy a necessary condition: two of x, y, z must be odd; the third even. To see why this is true, consider the possibilities: Either all three x, y, z are odd, or all three are even, or two of them are even, the third odd; or two amond them are odd, the third even. The first possibility is ruled out by considering () modulo : The left-hand said would be odd, while n is even. The second possibility is ruled out by considering () modulo 6: The left-hand side would be congruent to zero modulo 6; while n8n 8(mod 6), since N is odd The third possibility is also ruled out modulo ; the left-handed side would be odd, while n is even. We conclude that if (x, y, z) is a solution to (); then two among x, y, z must be odd, the third even. Also, due to the symmetry of (); if x0, y0, z 0 is a solution to (); so are the other five permutations of this triple: the triplesx, z, y, y, x, z, y, z, x, z, x, y, and z we may assume, (x and y odd, z even) in equation (). By Lemma, equation () is equivalent to (); x y xy z x y xy z 8N (3), y, x, Therefore Since x y(mod ) and z 0(mod ). Both factors on the left-hand side of equation (3), are even. But N is odd. Therefore (3) implies Either Or alternatively, x y xy z 4d x y xy z d (4a) x y xy z d x y xy z 4d (4b) 5

6 Where d, dare integers such that, dd N The integers d, d (both positive or both negative) are odd integers, divisors of N; and whose product is N. If (4a) holds, then by subtracting the second equation from the first we obtain, 4xy 4d d ; ( xy d ) d, which is impossible since d is odd. The same type of contradiction arises in the case of (4b). The proof, by contradiction is complete. Theorem Let n be a positive integer, and a, b, c, d integers such that a d b c;. Then the rational triple and abcd n b c, a b, a c is a solution to the equation x y z x y y z z x n And therefore (due to symmetry and even exponents), the triples ( xo, yo, zo) ; are also rational b c a b a c solutions; where (xo,yo,zo) is a permulation of,, ; and with any of the eight possible combinations allowed. Proof b c a b a c Since a d b c, it follows by Lemma 3(i) that the triple ( x, y, z),, is the unique solution to system of equations, x y z a x y z b x y z c x y z d Therefore, by multiplying the four equations memberwisel it is clear that the same rational triple is a solution to the equation, ( x y z)( x y z)( x y z)( x y z) abcd n ; which is equivalent, by Lemma, to equation (). Therefore the above rational triple is a solution to x y z x y y z z x n In the case of n 4 43; we have 4 3. So, with a 4, d, b 3, c. We obtain 5 the rational solution ( x, y, z),,. 6

7 4. A remark Remark Since equation () is symmetric with respect to the three variables x, y, z, and since all the exponents of the variables are even; it follows that if (xo, yo, zo) is a solution to equation (); then so are the following forty eight (not necessarily distinct) triples (with (xo, yo, zo) being among them): xo yo zo xo zo yo yo xo zo yo, zo, xo, zo, xo, yo, zo yo, xo ;,,,,,,,,, With all eight sign combinations being possible in each of the six permutations. Observe that if all three absolute values x0, y0, z0 are distinct positive integers or rationals. Then the above 48 triples are distinct. If x0, y0, z0 are positive with two of them being equal; the third different; then only twenty four of these triples are distinct. Likewise if one of x0, y0, z0 is zero; the other two distinct positive integers. If one of x0, y0, z0 is zero and the other two equal positive integers; then only twelve of the above 48 triples are distinct. If two among x0, y0, z0 are zero; the third a positive integer, then only 6 are distinct. However, this last combination can occur only when n is the fourth power of a positive integer. Theorem 3 and its proof Theorem 3 Let n be a positive integer. Then the 3-variable equation, x y z x y y z z x n ; has an integer solution if, and only if, there exist positive integers a, b, c, d, which satisfy the three conditions a d b c, abcd n, and a b c d mod (i.e. all four are even or odd) Proof First, assume that a, b, c, d are positive integers satisfying the above three conditions. Then, the three rational numbers, b c, y a b o o, a x z c o ; are all integers since a, b, c are either all even; or are all three odd (by the third condition). Moreover, since a+d=b+c, it follows by Lemma 3(i), that the triple is x, y, z, the unique solution to the system,

8 x y z a x y z b ; which in turn implies that x0, y 0, z 0, is an integer solution of the equation, x y z c x y z d x y zx y zx y zx y z abcd n; and thus, by Lemma, a solution of equation (). Next we prove the converse. Suppose that x0, y 0, z 0, is an integer solution to equation (). Then, by Lemma it follows that x, y, z, is a solution of equation (3); and so, x y z x y z x y z x y z n (5) o o o o o o o o o o o o By (5) it follows that xo yo zo d xo yo zo d xo yo zo d3 xo yo zo d 4 (6a) And with the integers d, d, d3, d 4 satisfying, ddd3d4 n (6b) According to (6a), the triple x, y, z, is a solution to the system x y z d, x y z d, x y z d, x y z d 3 4 Therefore, by Lemma 3 it follows that, d d d d o, o, d x y z d o 3 3 (7a) And also that, d d4 d d3 (7b) Since xo, yo, zo are integers; it follows from (7a); that all three integers d, d, d3 must have the same parity; they must all be even or all odd; and so by (7b), d4 must also have the same parity. Therefore, d d d d (mod ) (8) 3 4 Since n is a positive integer, then (6b) and (7b) we deduce that one of three possibilities must occur: Possibility : All four d, d, d3, d4 are positive integers. Possibility : All four are negative integers. 8

9 Possibility 3: One of d, d4is positive, the other negative. And likewise, one of d, d3is positive, the other negative. If Possibility holds, we are done since (6b), (7b), and (8) are the three conditions we seek to establish. If Possibility holds; by setting a d, b d, c d3, d d4. Then (6b), (7b), and (8);imply abcd=n, a+d=b+c, and also a b c d(mod). We are done in this case. If Possibility 3 occurs. Then there are four possible combinations, all treated similarly. So, when d 0, d4 0, d 0, d3 0. Just set a d, c d4, b d, d d3; and we are done: (6b), (7b), and (8) imply, abcd n, a d b c, a b c d(mod) 5. An Application of Theorem 3: Theorem 4 Theorem 4 Let n be a positive integer, n=p, a prime; or n=4; or n=pq, a product of two distinct primes p and q. Then, the equation Proof x y z x y y z z x n has no integer solutions. According to Th.3, this equation will have an integer solution, if and only if there exist positive integers a d b c, abcd n, a,b,c,d satisfying the three conditions and a b c d(mod) If n=p, a prime. Then abcd=n=p implies that three of the positive integers a, b, c, d; must equal, the fourth 0. But then the additive condition a d b c cannot be satisfied, since one side will equal ; the other side p 3. If n pq ; abcd pq implies that either one of a, b, c, d is pq, the other three ; which renders a+d=b+c impossible, since one side will be pq 3 7 other side. Or alternatively, one of a, b, c, d is p, another is q ; the remaining two, each being. And so a d b c implies either p q; or =p+q, both impossible since p, q are distinct primes. Finally if, n 4, a quick check shows that the first two conditions are satisfied only when one of a, d is, the other. And likewise, one of b, c is, the other ; so and 4. However, the congruence condition obviously fails. We have shown that there exist no positive integers a, b, c, d satisfying all three conditions. Thus equation () has no integer solutions. 6. Theorem 5 and it s proof The following theorem provides a method for finding all the nonnegative integer solutions of equation (). Theorem 5 9

10 Let n be a positive integer and consider the equation, x y z x y y z z x n Then, if x, y, z, is a nonnegative integer solution to this equation, then one of three sets of conditions must occur: Either b c a b a c xo, y0, z0, where a, b, c, d are positive integers C : satisfying a d b c, abcd n, a b c d(mod ); and a b, a c, a d Or Or b c a b a c x0, y0, z0, z where a, b, c, d are positive integers C : satisfying a d b c, abcd n, a b c d(mod ); and abc. c b a b a c x0, y0, z0 where a, b, c, d are positive integers C3 : satisfying a d b c, abcd n, a b c d(mod ); and a c b Conversely, every nonnegative solution can be obtained from one of the three sets of conditions; C, C, or C 3 Proof First we prove the converse. Suppose that one of three sets of conditions is satisfied. If C is satisfied, then it follows that x0, y 0, z0 are nonnegative integers, as a b c(mod) and a b, a c clearly imply. Also, since a d b c, Lemma3(i) implies that x, y, z is the unique solution to the system, x y z a, x y z b, x y z c, x y z d. And, since abcd n, by Lemma, it follows that x0, y 0, z0 is a nonnegative integer solution of equation ()

11 If C is satisfied, a similar argument shows that x0, y 0, z0 is a nonnegative integer solution to (). Lemma 3(i) is applied with A=a, B=b, C=-c, and D=-d; which gives ( x y z)( x y z)( x y z) ( x y z) ab( c) ( d) abcd n If C is satisfied, we apply Lemma 3(i) with A=a, B=-b, C=c and D=-d; a( b) c( d) abcd n. Again x 0, y 0, z0 is a nonnegative solution of equation () Next, suppose that x 0, y 0, z0 is a nonnegative solution of equation (); and therefore by Lemma, of equation (3) as well: +y + z -y + z +y - z -y - z x x x x n (9) x 0, y 0, z 0; we have x y z 0; in fact, Since a x0 y0 z0 0; with x0 0, y0 0, z0 0 a x0 y0 z0 ; since not all nonnegative integers x0, y0, z0 can be zero; by (9) and the fact that n is a positive integer. (0) It follows from (9) and (0) that either all three integers -y + z, +y - z, -y - z positive. Or otherwise, two of them must be negative; the third positive. If all three -y + z, +y - z, -y - z x x x ; are x x x are positive integers. Then, x0 y0 z0 a, x0 y0 z0 b, x0 y0 z0 c, x0 y0 z0 d with a, b, c, d positive integers (0a) Clearly, since x0, y 0, z0 are nonnegative; Also from (0a) implies a b, and a c. Also from (0a) and (9), we have abcd=n. Moreover (0a) says that x 0, y 0, z0 is a solution to the linear system, x y z a, x y z b, x y c c, x y z d Therefore by Lemma 3 it follows that b c a b a c a+d=b+c; and that x0, y0, z0 ; which in turn implies that, since x0, y0, z0 are integers; we must have abc(mod); and by a+d=b+c; a b c d(mod); It is now clear that all the conditions in C are satisfied. If two of the factors x0 y0 z0, x0 y0 z0, x0 y0 z0; are negative, the third positive; there are three possibilities. The possibility x0 y0 z0 0, x0 y0 z0 0, x0 y0 z0 0; is ruled out; since x y z 0 and x y z 0 imply x <0 contrary to x Of the other two possibilities: x y z 0, x y z 0, x y z 0. If

12 We then have, x0 y0 z0 a, x0 y0 z0 b, x0 y0 z0 c, x0 y0 z0 d with a, b, c, d positive integers. (0b) Since y0 0, it follows from the first two equations in (0b) that a b. And since x 0 0, it follows from the second and third equations in (0b) that b c; so that a b c. From (0b) and (9) we deduce that, ab( c) ( d) n; abcd n Next, apply Lemma 3 with A=a, B=b, C=-c, D=-d. It follows that b c a b a c x0, y0, z0, and A D B C; a d b c. Also, since x0, y0, z0 are integers; it follows that a b c(mod); and from a d b c; we further have a b c d(mod ). Lastly, the remaining possibly is treated similarly as the previous one. x y z 0, x y z 0, x y z 0. We have, If x0 y0 z0 a, x0 y0 z0 b, x0 y0 z0 c, x0 y0 z0 d where a, b, c, d are positive integers It follows that since z 0 a c; and that c b (since z0 0); And so a c b We also have a( b) c( d) abcd n After that, we apply Lemma3 with, A=a, B=-b, C=c, D=-d. We omit the details An application of Theorem 5: Theorem 6 Consider the case n p, p an odd prime. We can use Theorem 5 to determine all the nonnegative integer solutions of equation (). Observe that if four positive integers a,b,c,d satisfy the conditions, abdc p and a d b d Then since p is prime; one of a, d must equal p, the other ; and one of b, c must equal p, the other. And since p is odd, all four a, d, b, c are odd, so the congruence condition a b c d(mod) is satisfied. Also note that the possibility that one of a, b, c, d is by the condition a+d=b+c. p ; the other three equal to ; is obviously ruled out

13 We know from Theorem 5, that every nonnegative solution can be obtained from one of the sets of C C C. Also note that in all three C, C, C ; a max a, b, c conditions,, or 3 3. This clearly implies that always, a=p. Therefore: If C is satisfied, then a=p, d=; and either b=p, c=; or b= and c=p. We obtain two nonnegative p p p p solutions:,,0,0,. If Cis satisfied, we have ab c. Therefore, a p, d, b p, c. And therefor according to C, we obtain the nonnegative solution p p,0,. If C3 is satisfied, we have a c b. Thus, a=p, d=a, b=, c=p. We obtain the nonnegative solution p p,,0. From these four nonnegative integer solutions, all integer solutions of equation () can be obtained from permutations and combinations of signs. We have the following. Theorem 6 Let n p, p an odd prime. Then, the 3-variable equation x y z x y y z z x n, has exactly twenty four integer solutions: e f e f e f f e f e f e,,0,,0,, 0,,, 0,,,,,0,,0, ; p p where e, f. And in each triple, all four sign combinations can hold. 8. When n is a perfect square: n k In this section, we determine all the integer solutions, with xyz=0, in the case n k, k a positive integer. We also determine, when n k, all the rational solutions of equation (), with xyz=0. To accomplish our goal we need three results. Result Let k be a positive integer k Consider the -variable equation, u v k (i) If k (mod 4), then this equation has no positive integer solutions. (ii) If k 0,, or3(mod4). Then this equation has positive integer solutions. The solution set, in positive 3

14 integers u and v can be parametrically described as follows: d d d d u, v ; where d, d are positive integers such that d d, d d(mod ), and dd k Each positive integer solution pair (u,v) can be obtained once in this way: if D, Dare positive integers such that D D, D D (mod), D D k; and with d D or d D. Then the integer solution D D D D pair (U,V) is distinct from the solution pair (u,v); where U and V. Proof (i) An integer square can only be congruent to 0 or modulo 4; according to whether that integer is even or odd. Therefore, u u u v, 0,0, or 0 0(mod 4); v 0,, ; or 0(mod 4); v Therefore u 0,, or 3(mod 4). v cannot be congruent to mod4; and so there are no integer solutions. (ii) If k 0(mod 4) then k can be written as a product of two positive even integers: k u v u v u v ( )( ); and so we must have, u v d u v d u v u v k d, (, since,,, and ) Where d, and dd k. From which we obtain, d d, d u v d d, d are even positive integers with If k or 3(mod4); then k is an odd integer; k d d, with d d (mod), d d. Again, d d d u d the details., v. That each solution is obtained in this way only once is fairly obvious; we omit The following result states the well-known parametric formulas that describe all the positive integer solutions to the Pythagorean equation. This result can be found in number theory books, in wikipedia on the world wide web, and other internet sources. For a wealth of information regarding Pythagorean triples and triangles, an excellent source can be found in reference []. Also, in the same reference, a detailed derivation of the parametric formulas stated below. 4

15 Result All the positive integer solutions to the 3-variable equation as follows: u d( k k ), v d ( k k ), w d( k k ) ; or alternatively, u d k k v d k k w d k k ( ), ( ), ( ). Where 5 u v w can be parametrically described d, k, k are positive integers such that k k, k k (mod ) (i.e. one of k, kis odd, the other even); and gcd( k, k) (i.e. k and k are relatively prime. In a 006 paper published in the journal Mathematics and Computer Education (see reference [3]), By this author; the reader can find parametric formulas describing the entire set of positive integer solutions to the 3-variable equation u kv w ; where k, is a fixed positive integer. In the same article, the reader can find a step-by-step derivation of all the positive integer solutions to this equation. We state this result. Result 3 Let k be a positive integer, k. The entire set of positive integer solutions of the 3-variable equation u k v w, can be parametrically described as follows: d k m k m d ( k m k m ) u, v dmm, w Where d, k, k, m, m are positive integers such that k k k, gcd( m, m ), (i.e. m and m are relatively prime) and dk m dk m Consider the case where two among x, y, z in equation (), are zero; say y=z=0. Then, 4 x k x k ; which shows that if k is a perfect square; k m ; with m, (mod ) then x m, and so there are integer solutions; otherwise there are no integer solutions. What about rational solutions? Recall that the square root of a positive integer is either a positive integer; or otherwise an irrational number. This statement is equivalent to the following statement: A positive integer is equal to the square of a rational number; if, and only if, that same integer is a perfect or integer square. This holds true for any natural number exponent, and not just squares. This statement can be fairly easily proven. The interested reader may refer to reference []. Theorem 7 Let n k, k a positive integer, and consider the equation. Let S and Sbe the sets, x y z x y x y y z z x n S {( a, b, c) ( a, b, c) is an integer solution () with two among a,b,c being zero} ; and S a, b, c a, b, c is a rational solution of equation (), with two among a, b, c being zero

16 Then, (i) If K is not a perfect or integer square, S S ; equation () has no such integer or rational solutions. (ii) If k m, m a positive integer; then, S S m,0,0, 0, m,0, 0,0, m There are six integer and six identical rational solutions in this case. Next, we consider the case where exactly one of the three variables is zero; the other two nonzero. If we take z=0, and 0 xy. Then x y k x y k k ; ; with. And because of the symmetry of equation (), to find all such integer solutions; it suffices to find all the positive integer solutions of the equation, x y k. But Result, provides a complete answer to this question. Accordingly, we can state Theorem 8. Theorem 8 Let () n k, k an integer with k. Consider the equation, x y z x y y z z x n And let S be the set, S a, b, c a, b, c is an integer solution of equation (), with exactly one of a,b,c being zero (i) If k (mod4), S ; there are no such solutions. (ii) If k 0,, or 3(mod 4); then the set S is a finite nonempty set which can be described/determined in the following way: For each pair of divisors d and d of k; such that d d(mod ), d d and dd k; define the d d d d positive integers e and f (The integers d and d satisfy the conditions of Result ). From a specific such pair ( d, d) of divisors d and d of k; there are exactly twenty four distinct integer solutions (with exactly one of x,y,z being zero) to equation (): e, f,0, e,0, f, f, e,0, f,0, e, 0, e, f, 0, f, e. The number of integer solutions in the set S is 4N; where N is the number of the divisor pairs ( d, d) that satisfy the conditions, d d(mod ), d d, and dd k (it is clear from Result, that the twenty four such solutions generated by another divisor pair D D will be distinct from the twenty four solutions generated by the pair ( d, d )., Now, let us look at the case of rational solutions (with exactly one of the variables being zero) of equation (). If we set z=0, we have x y k ; and due to symmetry in (), it suffices to find the positive 6

17 rational solutions of There we obtain, w u x y k. By putting, x, y ; v v where w, v, and u are positive integers. u k v w, which is the equation of Result 3. Accordingly we get, w m m x k k v m m ; and y u k m k m ; where v m m relatively prime positive integers. Theorem 9 Let () Let n k, k an integer with k k, k are positive integers such that kk k. And m, m are. Consider the equation, x y z x y y z z x n S a, b, c a, b, c is a rational solution to equation (), with exactly one of a, b, c being zero The set S is an infinite set which is the union of six classes of solutions; which can be parametrically be described as follows: Let k, kbe positive integers such that kk k And m, m (arbitrary) relatively prime positive integers. k m m k m m Also, e note that e 0 f k m m k m m And also with k m k m (so that f 0) Class C e, f,0 Class C e,0, f Class C 3 f, e,0 Class C 4 f,0, e Class C 5 0, e, f 7

18 Class C 6 0, f, e S CUC UC UC UC UC Then, (With all sign combinations being possible) We finish this section with the case n=. The case of integer solutions with xyz=0 is immediate:,0,0, 0,,0, 0,0,, are the only integer solutions with two of the variables being zero. The same is true for such rational solutions. When only one of the variables is zero; there are no integer solitions since (as it is easily verified), the equation x y has no positive integer solutions. The w u same equation though, has positive rational solutions. By setting, x, y ; v v where w, v, u are positive integers. We obtain w v u. And from Result we further get, w k k u k k x, y ; or alternatively, v k k v k k k k x k k k k k k, y Where k, kare relatively prime positive integers such that k k and k k (mod ) We have the following theorem. Theorem 0 Consider the equation, x y z x y y z z x. (a) This equation has six integer solutions with two of the variables being zero:,0,0, 0,,0, and 0,0,. These are also the only such rational solutions. (b) The above equation has no integer solutions with only one of the variables being zero. (c) The set S of rational solutions with only one of the variables being zero; is the union of twelve classes of solutions which can be described as follows: Let k, kbe positive integers such that k k, k k (mod ); and k, k are relatively prime. k k k k k k kk Let e, f, g, and h. k k k k k k k k (Clearly all four; e, f, g, and h; are positive). Class C e, f,0 Class C e,0, f Class C 3 f, e,0 8

19 Class C 4 f,0, e Class C 5 0, e, f Class C 6 0, f, e For Classes C7 throughc ; just replace e by g ; and f by h; and repeat the process. (with all sign combinations being possible) Remark Note that Th. 9, there is the condition k m integer. One may ask the question: when is k m, k m k m ; to ensure that f is not zero; and so a positive or equivalently, k m? Since m and m k m are relatively prime; that would imply k m and k m, for some positive integer. And since kk k; this in turn would imply m m which can only occur when k is a perfect square, k; k N, N a positive integer. And so in such a case, mm N; so the product of the two positive integer parameters m and m, would be a divisor of N. 9. The equation 4 4 x y z xy xz y z n If we replace x by x in equation (), we obtain an equation which is quadratic in x : x y z x y 4 z 4 y z n ; or equivalently x y z x y z n 0 () Equation () is symmetric only with respect to the variables y and z. Applying the quadratic formula x y z yz n () yields, 0. The equation Integer Solutions 4 4 x y z xy xz y z n : 9

20 It is clear from () that equation () will have integer solutions if and only if the integer yz an integer square: yz n T where T is a nonnegative integer. But since n (3) and ( yz) 0; it is clear that T must be a positive integer The following analysis will quickly lead us in formulating Theorem. ) By inspection, by looking at equation (); we see that if the positive integer n is not a perfect square; the () cannot have any integer solutions with xyz 0. Indeed, if x 0 in (); then n y z. And if y 0or z 0 ; then n x. ) If n is a perfect square; a) Solutions with 0 n k ; k a positive integer. x require that, to y and z; it suffices to determine all the integer solutions of If 0 n is y z k ; and because of the symmetry with respect y z k k m, m a positive integer. Then the solutions with z 0; have y m. If k is not a perfect square; then there are no solutions with yz 0. If k (mod 4), there are no integer solutions. If k 0,, or 3(mod 4); the integer solutions of positive integer solutions of with yz 0; are determined by the y z k; y z k; just apply Result. b) Solutions with x 0 and yz 0. From (3), we get T k. And so for yz0, we have (from()) xk; and for y0, z0we obtain exception that if 3) Solutions with yz 0 k m ; y m, if the minus sign holds in x y k; where y can be integer; with the x y k x ; since 0. As we have seen in # above; unless n is a perfect square; x 0 as well. If n is a perfect square, a particular combination of y and z may result in x 0. Regardless, solutions with yz 0 ; can be completely determined from the positive integer solutions of equation (3): T yz n According to Result, if n (mod 4), there are no positive integer solutions. Otherwise, we must have d d, d T yz d ; where d and dare positive integers such that, d d, d d (mod), dd n. The second equation above states that, 4 yz dd; which implies d d (mod 4) If n 0(mod 4), n the combination d, d ; ensures that this equation is always satisfied for some divisors

21 d and d of n. Likewise, when n (mod 4); one such combination is d n and d ; so that d d n 0(mod 4) (When n is prime, this is the only such combination) However, for n 3(mod 4), no such d and d can exist. Indeed, dd 0(mod 4) and dd n 3(mod 4) And so, d d (mod 4) which implies, d d d d ( d ) n 3(mod 4); which is impossible since in fact d (mod 4), by virtue of the fact that d is odd. Thus, equation () has no integer solutions with yz 0 if n 3(mod4). Theorem Let n be a positive integer. Consider the equation 4 4 x y z xy xz y z n ) If n is not a perfect square, then the above equation has no integer solutions with xyz 0; that is, with at least one of x,y, or z being zero. ) Suppose that n k, k a positive integer. a) If k is not a perfect square. Then the above equation has no integer solutions with x=0; and with one of y, z also being zero. b) If k m, m a positive integer; then the integer solutions with x=0; and one of y. z also zero are: (0, m,0), (0,0, m); four such solutions in total. c) If k (mod 4), there are no integer solutions with x=0. d) The integer solutions with x 0 and yz=0 can be described as follows: x t k y t z,, 0; being one family of such solutions. And x t k y z t, 0, ; being the other. Where t can be any integer. And with one exception: t m; t m; so that x 0 If k m, m ; and the minus sign holds; then 3) The integer solutions with yz 0can be described as follows: If n or 3(mod4), then there are no integer solutions. If n 0 or (mod4); then all such integer solutions can be described in the following way. For every pair of divisors d and dsuch that n dd n, d d, d d(mod4) (when n 0(mod 4); d, d is such a combination. For n (mod 4); d n, d is such a combination); define d d and d e f d Then all 4 the integer solutions determined by this pair of divisors are given by x e f,

22 e y, z= ; where is a positive integer divisor of the positive integer e; and with all sign combinations being possible. The following theorem is the immediate consequence of parts and 3 of Theorem. Theorem Let n be a positive integer with 4 4 x y z xy xz y z n n or 3(mod4). Then the 3-variable equation, has no integer solutions.. The equation 4 4 x y z xy xz y z n : Rational solutions As we have already seen this equation is equivalent to (), whose solutions are given by (): x y z ( yz) n First, consider rational solutions with yz 0. Clearly, such solutions exist precisely when n is a rational square; and since n is a positive integer, n is a rational only if it is a perfect or integer square. So, if n k, k a positive integer; all the rational solutions with yz=0, are given by ( k,0,0), ( r k, r,0), ( r k,0, r); where r is arbitrary positive rational; and will all the sign combinations being possible. u Consider rational solutions with yz 0. If we put yz, where u and v are positive integers. Then, v u nv u nv ( yz) n, which will be rational if and only if, v v positive integer w. By result3, we must have u nv w, for some d nt nt u, v dt t, and d nt nt w ; where n, n, t, t are positive integers such that nn n, gcd( t, t), and dnt dnt (mod ) And with nt nt (so that u 0) (4)

23 If we take y r, a positive rational. Then from yz z n t n t 4r t t. Therefore from () we further get, nt nt nt nt x r 6r t t t t Theorem 3 Let n be a positive integer. Consider the equation u u we get z ; and thus by (4) we have, v rv x y z w v ; and by (4) 4 4 x y z xy xz y z n (a) If n is not a perfect square, this equation has no rational solutions with yz=0. (b) If n k, k a positive integer, then all the rational solutions with yz=0 are triples of the form: ( k,0,0), ( r k, r,0), and ( r k,0, r); where r is an arbitrary positive rational; and with any sign combination being possible. (c) All the rational solutions with yz 0, can be parametrically described as follows: nt nt nt nt x r 6r t t t t n t n t y r, and z 4r t t Where r is an arbitrary positive rational, any sign combination being possible; and with n, t, n, t being positive integers such that nn n and gcd( t, t) And with n t References n t. [] Crux Mathematicorum, No 5 (May 03), Vol 38, Contest Corner problem CC4, page 74 Given the equation, x y z y z z x x y 4 (a) Prove that the equation has no integer solutions (b) Does this equation have rational solutions? If yes, give an example. If no, prove it [] W. Sierpinski, Elementary Theory of Numbers, Warsaw 964, 480pp. ISBN: For the result that states that is a natural number is the mth power of rational number, and m is natural; see Theorem 7, on page 6 For the Pythagorean equation [3] Konstantine Zelator, The Diophantine equation x y z, see pages 38-4 x ky z, and Integral Triangles with a Cosine Value of n Mathematics and Computer Education Vol 30, No 3, pp 9-97, Fall 006 3

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