Certain Properties of Pythagorean Triangles involving the interior diameter 2ρ, and the exterior diameters 2 ρα,2 ρβ,2ρ. Part II: The legs case

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1 Certain Properties of Pythagorean Triangles involving the interior diameter ρ, and the eterior diameters ρα, ρβ,ρ Part II: The legs ase Konstantine Hermes Zelator Department of Mathematis College of Arts and Sienes Mail Stop 94 The University of Toledo Toledo, OH USA

2 1. Introdution Given three positive integers s α, β,, ongruent triangles will eist on the two- dimensional Eulidean plane, with side lengths α, β, ; as long as the three integers satisfy the triangle inequalities α β >, α > β, β > α. And depending on the type of integers α, β, involved, one may be able to onstrut suh a representative triangle (of the said family of ongruent triangles) by using ruler and ompass; or some other means. Even if suh a triangle is not onstrutible (by using traditional instruments suh as a ruler and a ompass), it will ertainly eist in a mathematial sense. It is also easy to see that there eist infinitely many integer-sided triangles with one side length being a perfet or integer square; and one of the four positive integers α β (perimeter), α β, α β, α β being also and integer square. Indeed, if we take α=k for some positive integer k, and say, α β=, where is also a positive integer. Then, the numbers β and an be hosen so that, β = k (with > k) > k,whih in turn requires that the positive integers and k satisfy > k > k ), and κ < β < κ or equivalently, β <κ. For eah value of t = κ 1, k,...,0,1,,..., κ, κ 1, a triangle with integer sidelengths k t k t α= k, β=, =, is generated; sine, as it is easily seen, the three triangle inequalities are satisfied. The four real numbers α β, α β, α β, α β, are also reognizable as being fators in the under the radial quantity in Heron s formula for the area of a triangle. As we will see in Setion of this paper, if we divide twie the area of a triangle by eah of the numbers α β, α β, α β, α β, we obtain respetively

3 ρ, ρ ρ, ρ. These are the diameters of four important irles: the triangle s inner or α β insribed irle and the three eterior irles, eah of whih is tangential to one side of the triangle and tangential to the two straight lines ontaining the other two sides (but not tangential to the other two sides themselves). Eah of the three enters (of these three irles) is the intersetion of one internal bisetor (i.e. bisetor of one of the triangles internal angle) and two eternal bisetors (i.e. bisetors of two eternal triangle angles). In setion 3, we easily show that if a triangle ΑΒΓ is a right one with the 90 o degree angle at the verte A; then, in fat, ρ=β α, ρ =αβ, ρ =αβ,ρ =α β α β (so in this ase, α is the length of the hypotenuse). Whih brings us to Setion 4, whih is the fous of this paper, namely, Pythagorean triangles. Therein we obtain simple formulas for ρ, ρα ρβ, ρ, in terms of the two integer parameters m and n that generate the family of primitive Pythagorean Triangles. In Setion 5, we state seven well-known results from number theory. Results 3-7 an be found in standard tets and books of number theory we (we offer referenes [] and [3] for this); the relevant material overed in the first half of a first ourse in elementary number theory. In Result 1, we state the parametri formulas whih desribe the entire family of solutions of the diophantine equation y =z, with (,y)=1. This is well-known and an be found in E.L. Dikson s landmark book (see [1]). It an also be found in [4]. In Result, we state the parametri formulas that desribe the entire family of solutions to the diophantine equation y =z, with (,y)=1. More on this in the last setion of this paper, Setion 8. 3

4 In Setion 6, we establish Proposition 1: there eist infinitely many primitive Pythagorean triangles with one leg being an integer square. This is a well-known result, but we need to have it at our disposal for what it follows later. This then brings to the entral question of this paper. Whih is, what an be said about Pythagorean triangles whih have one sidelength being an integer square, and one of the four integers ρ, ρα ρβ, ρ, also a square? In Setion 7, we prove Theorem 1: that in a primitive Pythagorean triangle, with a hypotenuse length α ; β the even leg length, and the odd leg length; eah of the following two ombinations is impossible: 1) a square and ρ also a square. ) a square and ρ also a square. α Note that if in a primitive Pythagorean triangle we require that one leg has length an integer square; and that one of the four diameters ρ, ρα ρβ, ρ is also an integer square. Then there are eatly eight ombinations. Aording to the above, two of these ombinations are not possible. The remaining ombinations are: 3) β a square and ρ also a square. 4) β a square and ρ α also a square. 5) a square and ρ also a square. 6) 7) β β a square and ρ a square and ρ β also a square. also a square. 8) a square and ρ β also a square. In Theorem, also in Setion 7, we show that all of the si ombinations are possible, and we parametrially desribe eah of the si families of suh primitive Pythagorean triangles. We also present 1 numerial eamples. 4

5 We lose this paper with Setion 8, in whih we inlude some historial omments about the diophanline equation y =z, (,y)=1; as well as a brief sketh of a derivation of the general solution of this equation. Notation 1. We will use the standard notation (a,b) to denote the greatest ommon divisor of two integers a and b.. To denote a partiular solution to a diophantine equation we will use the braket notation. For eample, { 0, y 0, z 0 }, an stand for a solution to a three-variable diophantine equation in, y, and z. 3. The symbol Z will denote the set of integers; Z the set of positive integers. 4. The notation a b, will mean that the integer a is a divisor of the integer b. The same meaning an be onveyed by using the phrases a divides b ; or b is divisible by a ; or a eatly divides b ; or the language of ongruenes, b 0(mod a). 5. The notation AB will mean line segment AB, and AB will denote the length of that line segment. 5

6 . Formulas for the diameters ρ, ρ ρ, ρ in terms of the sidelengths α, β,. From Figure 1, we easily see that area (of triangle) area area Also, we have Δ ABI Δ BIΓ Δ BIΓ = 1 ρ 1 = ρ α } 1 = ρ β { y = y z = α z = β α β Δ area (of triangle) 1 ABI = ρ ( α β ). } (algebraially solving) { } β α = α β y = α β z = Net, AM = AK (sine M and K are points of tangeny of straightlines AM and AK, with the irle of enter ); whih gives BM = β ΓK. But also, BG = BM and ΓG O α = ΓK ; thus sine BG ΓG = α ; it follows that BM ΓK = α. Combining this β α with the above BM - ΓK =β ; β α; an easy solving produes BM = and ΓK = Δ ρα ρα Moreover from right triangle Ο we see that, tan θ= =, in view of α MA AM ( αβ ) ρ tan θ =. βα αβ AM = BM = =. and from right triangle ΑΔΙ, we also have α β ρα ρ Therefore we obtain, =. ( αβ/ Using the fat β α (AREA) = and ρ= ; αβ where AREA simply stands for the area of triangle AB Δ Γ we finally arrive at ρ α = 4(AREA) α β by yliity or permutation of the three letters A BΔ Γ, we an state the formulas for the four dameters: 6

7 4(AREA) ρ =, α β ρ α 4(AREA) =, β α ρ β 4(AREA) =, α β ρ 4(AREA) = α β (1) Figure 1 AB =, B Γ =α, Γ A =β AΔ= = AΕ BΔ= y= BZ Γ Z = z= ΓE ΔΙ = Ι Z = IE = ρ M O α = ρ α 3. The Case of Right Triangles If Δ ABΓ is a right triangle with the ninety degree angle at the verte A; then α beomes the hypotenuse length and β and the two leg lengths. We have, α = β β= ( βα)( β α), whih when ombined with first two formulas in (1); and the fat that 4(AREA)= β ; it yields ρ = β α and ρ α = α β Note that ρ α is equal to the perimeter of the right triangle. Similarly, in view of the algebrai equivalene β = ( α β )[ α ( β )]; ouhed with (again) 4(AREA) = β, and the last two formulas in (1) ; we obtain, ρ = α β and ρ = α β. β Altogether, 7

8 ρ = β α, ρ = α β, ρ = α β, ρ = α β () α β 4. The Case of Pythagorean Triangles The entire family of Pythagorean Triangles or triples ( α, β, ) an be desribed by the wellknown parametri formulas, α=δ ( ), β=δ(mn), =δ( ) (3) m n m n where m,n, δ are positive integers suh that (m,n)=1, m>n, and m n 1(mod ) (i.e. m and n have different parities, one is even, the other odd). This parametri desription an be found in most books of number theory; tetbooks or more sholarly books. For eample, you may refer to [1], [], or [3]. For δ =1, the Pythagorean triple or triangle is alled primitive; with hypotenuse length α = m n, leg of even length β = mn, and leg of odd length = m n ; with (m,n)=1, (4) m n 1,m,n Z,andm> n. Applying formulas () and (4) we see that, If ( α, β, ) is a primitive Pythagorean triple, the four diameters ρ, ρα,ρβ, ρ are given by the formulas, ρ = n(m n), ρ = m(n m), ρ = n(m n), ρ = m(m n) ; (5) α β where the positive integers m,n satisfy the onditions in (4). 5. Seven Results from Number Theory Result1: The entire set of solutions {,y,z}, in positive integers, to the diophantine equation 8

9 = y z, with (,y)=1, an be parametrially desribed by the formulas, = k λ,y= κλ,z= k λ, where the parameters k, λ, an take any positive integer values whih are relatively prime;, (k, λ ) = 1, k, λ Z and with k odd; k 1(mod ) (so that both and z are odd) Result 1 is well-known in the literature and an be found in [1]. Result: The entire set of solutions in positive integers, to the diophantine equation = y z, with (,y)=1, an be parametrially (up to symmetry; the equation is symmetri with respet to and y; if {a,b,} is a solution, so is {b,a,}) desribed by the formulas, = k λ,y= k λ,z = ; k λ k λ k λ where k, λ an be any positive integers suh that ( k, λ) =1, k λ 1(mod ) (so all three,y,z, in any solution with (,y)=1; must be odd). For historial ommentary on the diophantine equation of result as well as a brief sketh of the derivation of the general solution, go to Setion 11. Result 3: Let a, b Z, ab 0, and (a,b)=1. The following statements hold true: (i) If Z,anda bi, then a must be a divisor of. (ii) If p 1...k; where p is a prime and,..., integers; then p must be a divisor of at least one of the numbers,...,. 1 k 1 k (iii) If a b 1(mod), then (a b,a b) = and either a b 0anda b (mod 4); or (vie-versa) a b and a b 0(mod 4) (iv) If a b 1(mod ), then( a ± ab ± b, a b ) = 1 for any of the four ombinations of signs, and if a b 1(mod ), then( a ± ab ± b, a b ) =, for any sign ombination. Parts (iii)-(v) of Result 3, are typially assigned as straightforward eerises in the first half of an introdutory number theory ourse. Part (i) an be found in number theory books either listed as an eerise or as a theorem: see [] or [3]. And part (ii) is used in proving the Fundamental 9

10 Theorem of Arithmeti (unique fatorization of integers into prime powers). Again standard material found in number theory books. n Result 4: ε ( ) = = ( ) n n a,b, Z, a,b 1, and ab ; then a =,b =, for some 1, εz, with, = 1,and = This is a well-known result, you may refer to [] or [3]. Based on Result 4, the following result an be easily proven. Result 5: If p is a prime number, and ab = p n ; a,b, ε Z,( a,b) = 1; then there eist 1, εz n n n n suh that either a = p 1,b= or alternatively a = 1 and b p ; 1 = = and with ( ), = 1 and The proof of Result 5 is easy, if we make use of Results 4 and 3(ii), Indeed, by Result 3(ii), we know that p must divide at least one of a and b; say pa. We then have a=pd, for some d εz. 1 Aordingly, ab n n n = gives p pdb = p ; db =. Clearly (d,b)=1, sine d is a divisor of a and n n (a,b)=1. By Result 4 we must have d =, b =, with (, ) = 1and =. Thus, a = p andb =. The other possibility, pb, leads to n n 1 n n 1 p a = and b =. n Result 6: (i) If p is a prime, and a,b, Z, with (a,b) = 1 and pab = ; then either n 1 n; a = p 1 and n n n 1 n 1 p b = ;ora= andb = ; for some positive integers 1, suh that ( 1, ) = 1 and 1 = (ii) In partiular when n= in part (i), we must have either a = p andb = ; or, alternatively a = andb= p

11 We leave the proof of Result 6 to the reader as an eerise. Here is a sketh of this proof. Apply Result 3 (ii) to onlude that the prime p must divide. Then use this fat to dedue that must divide either a or b; by using the hypothesis(a,b)=1 together with the fat that prime power; and in onjuntion with Result 3 (i). Finally, apply Result 4 to onlude the proof. n n Result 7: If a,b Z and a b, then a b In words, this an be stated as follows: If the nth power of a positive integer divides the nth power of another positive integer; then the former positive integer must divide the latter. This is a well-known result, refer to [] or [3]. 6. Infinitely many primitive Pythagorean triangles with one leg length being a square. Let us remark at the outset, that a Pythagorean Triangle an have only one leg whose length is a square. P. Fermat was the first historially known mathematiian who s used the method of infinite desent. He used this method (see Setion 8) to show that the diophantine equation n 1 p n 1 p is a 4 4 y = z has no positive integer solutions. This is a well-known result whih an be found in many number theory books. An immediate orollary of this is the non-eistene of Pythagorean Triangles both of whose leg lengths are integer squares. As we see below, the proess of desribing all primitive Pythagorean Triangles with one leg length being a square, is a fairly simple one. Proposition 1 (i) There eist infinitely many primitive Pythagorean triangles or triples { αβ},,, with the even leg length β being a square. The entire family an be parametrially desribed by the formulas, 11

12 α=, β =mn, ; m n m t1 = m n, where m= t 1 and n= t ; or = =, and n= t, where t 1 and t an be any positive integers suh that ( t1, t ) = 1 and t odd in the first ase; t 1 odd in the seond, and also with t > t or t > t respetively. 1 1 (ii) There eist infinitely many primitive Pythagorean triangles or triples { αβ,, }, with the odd length being a square. The entire family an be parametrially desribed by the formulas, α=, mn m n β=, = m n ; where m= t1 t, n = tt 1, where t 1 and t an be any positive integers, suh that ( t1, t ) = 1, t 1 t = 1(mod ) ( t 1 and t have different parities), and t > t (this ensures that there is no repetition of triples { α, β, }) 1 Proof: (i) If αβ,, are given by the stated formulas, then an easy alulation shows that β is square and that { αβ,, } is a primitive Pythagorean triple. Now, onversely, suppose that { αβ,, } is a primitive Pythagorean triple with β a square. We must have α= m n, β= mn, = m n, and β= ; for some positive integers m, n,, suh that (m, n)=1, mn 1 (mod), and m>n. From mn= and Result 6 with p=, it follows that either m= t 1 and n= t or m= t 1 and n= t. In the first ase we have m= t, and n= t. Now, (m,n)=1 and mn 1(mod); whih easily implies that t is odd and ( t1, t ) = 1. And m>n implies t1 > t. In the seond ase, m= 1 t and n = t. Again by virtue of (m,n)=1, m>n, and mn 1(mod), it easily follows that t 1 must be odd, 1 ( t, t ) = 1, and t > t. 1 (ii) As with part (i), one diretion in the proof is almost immediate. In the other diretion, suppose that { αβ,, } is a primitive Pythagorean triple with the odd length being a square. We 1

13 must have, α= m n, β= mn, = m n, and = ; for some positive integers m,n suh that (m,n=1, mn 1 (mod) and m>n. Then, = m = n. Clearly (m,n)=1, easily implies from the last equation that also (m,)=1=(n,). In other words {m,n,} is a primitive Pythagorean triple with m being the hypotenuse length and hene odd; and also from mn 1(mod), we infer that n must be even. Consequently, m = t t,n = t t,= t t, for some t 1, t εz, suh that ( t1, t) = 1, t1 t 1(mod ),andt1> t Primitive Pythagorean triangles with one leg length being a square and one of the diameters ρ, ρα, ρβ,ρ also a square. If { αβ,, } is a primitive Pythagorean triple with m n, mn, m n,m,n Z,(m,n) 1,m n 1(mod), and m>n, then there are α= β= = ε = eatly eight ombinations (with one side length a square and one of the four diameters also a square.) Combination 1: β a square and ρ a square Combination : β a square and ρ α a square Combination 3: β a square and ρ β a square Combination 4: β a square and ρ a square Combination 5: a square and ρ a square Combination 6: a square and ρ α a square Combination 7: a square and ρ β a square Combination 8: a square and ρ a square. 13

14 As Theorem 1 below shows, the two ombinations 6 and 8, annot really our. On the other hand, aording to Theorem, the remaining ombinations do our; and in eah ase (ombination) the entire family of suh triples is parametrially desribed. Theorem 1 (i) There are no primitive Pythagorean Triangles with the odd length a square and with diameter ρ α (also the perimeter of the triangle) also a square. (ii) There are no primitive Pythagorean Triangles with the odd length a square, and the diameter ρ also a square. Proof (i) By Proposition 1(ii), if suh a triple { α, β, } eists, we must have α= m n, β= mn, = m n, and with t1 t t t1 t t1 t m =,n = t,,(, ) = 1, 1(mod), for some t, t εz 1 In addition, ρ α = K, for some positive integer K. By (5), we know that ρ α = m(m n). Therefore, ( t t )( t t t t ) = K ; whih is ontraditory sine, in view of t1 t 1(mod ), the left-hand side is ongruent to ; but the right-hand side an only be ongruent to 0 or 1 mod4. (ii) As in (ii), we must have α= m n, β= mn, = m n,m = t t,n = t t ; for some 1 1 positive integers t 1, t suh that ( t1, t ) = 1 and t 1 t 1(mod ). Additionally, ρ = N, for 14

15 some N εz. And from (5), ρ m(m n). Consequently, ( t t )( t t t t ) = N ; = ) ( t t )( t t = N, whih is impossible modulo 4, sine the left-hand side is ongruent to, while the right-hand side an only be 0 or 1 mod4. Theorem (i) The primitive Pythagorean triples { α, β, } with β (the even length) a square and the diameter ρ α (also the perimeter) also a square, are preisely the members of the following family: Family F 1: α= m n, β= mn, = m n,andwith m= t,n = t, t1 > t,where t= κ λ, t1= κλ, κ 1(mod), ( κλ, ) = 1, κλε, Z (ii) The primitive Pythagorean triples { α, β, } with β (the even length) a square and the diameter ρ β also a square, are preisely the members of the following family: Family F : α= m n, β= mn, = m n,andwith m = t1,n = t, t1 > t,where t1= κ λ, t= κλ, κ 1(mod), ( κλ, ) = 1, κλε, Z (iii) The primitive Pythagorean triples { α, β, } with β (the even length) a square and the diameter ρ also a square, are preisely the members of the following family: Family F 3: α= m n, β= mn, = m n,with m= t1,n = t, t1 > t;andwith t1= κ λ, t= κ kλ λ κ, λεz, κλ 1(mod),( κ, λ ) = 1 15

16 (iv) The primitive Pythagorean triples { α, β, } with (the odd length) a square and the inner diameter ρ (also the perimeter) also a square are preisely the members of the following family: Family F 4 : α= m n, β= mn, = m n,andwith m = t1 t,n = t1t;where t1= κ, t= λ, t1> t κ, λε,( κ, λ ) = 1, κλ 1(mod ) Z (v) The primitive Pythagorean triples { α, β, } with (the odd length) a square and the diameter ρ β also a square, are preisely the members of the family F 4 of part (iv) of this Theorem. This is beause (as it will be easily seen in the proof) when is square; then ρ is also a square in, and only if, ρ β is also a square. (vi) The primitive Pythagorean triples { α, β, } with β (the even length) a square and the inner diameter ρ also a square, are preisely the members of the following family: Family F 6 : α= m n, β= mn, = m n,andwith m = t1,n = t;where, t1= κ λ, t= κλ; κ, λεz,and suhthat κ 1(mod ) and( κ, λ ) = 1 and with t1 > t Proof 16

17 First, we point out that in eah part, one diretion in the proof is straightforward: if { αβ,, } is a member of the given family, then it is a primitive Pythagorean triple with the stated property. Below, in eah part, we prove the onverse. If { α, β, } is a primitive Pythagorean triple with the stated property, then it must be a member of the given family. (i) By Proposition 1(i) we know that we must have, α= m n, β= mn; and with either m= t,n = t,and t 1; or alternatively m = t,n = t,and t 1(mod). And in either ase, with ( t1, t ) = 1. By (5) we know that, m(m n) = L, for some L Z. Sine m and n have different parities, m n 1(mod). But obviously, L must be even; and so L O(mod 4). Thus, m(m n) 0(mod 4); whih implies that m is even, sine mn is odd. Therefore, only the first of the above two possibilities an our. Namely, m= t andn = t,( t, t ) = 1 and with t odd (and also with t1 > t ) We have 1 1 t1 t1 t t1) t1 m(m n) = L 4 i ( ) = L ( L L, by Result 7. We set L= ( t ) i L1, 1 for some positive integer L 1. Furthermore, 4 t ( t t ) = 4t L t t = L ; whih shows that { t, t, L 1} is a solution in positive integers to the diophantine equation 1 y = z. By Result 1 we must have, t = k λ, t = κλ, κ 1(mod),( κ, λ ) = 1, for some positive integers κλ,. 1 (ii) This derivation is very similar to the one in part (i). Indeed, sine by (5) we must have n(m n), ρ β = = L for some L Z ; a ongruene modulo 4 implies that sine mn is odd, n must be even. This ouhed with Proposition 1(i), yields α= m n, β= mn,m = t,n = t, with t 1 odd, ( t1, t ) = 1, and t1 > t. Then, from n(mn)= L we easily obtain 1 ) 1 = ( t i ( t t ) L and by applying Result 7, we end up with t t = L 1, for some L1 Z ; 1 17

18 whih shows that { t, t, L 1} is a solution in positive integers to the diophantine equation 1 y = z ; hene, t = κ λ, t = κλ, for some positive integers κ, λ suh that 1 ( κλ, ) = 1, and κ 1(mod ). (iii) Sine ρ m(m n) is a square and m-n is odd (by virtue of mn 1 (mod)), it = follows that m must be even. (Again arguing modulo 4). Combining this with Proposition 1 (i) gives m= t1,n = t, with t odd, t1 > t, and ( t1, t ) = 1. Then, from ρ =square we obtain, ) 1 = ( t i ( t t ) L, for some positive integer. By Result 7 it follows that t is a divisor of L; put L = ( t ) i L1, for some integer L 1. Therefore the last equation gives, t1 t= L1; t1 = t L1. Note that sine t is odd, so must be L 1; and t 1 must be odd as well, as a ongruene mod4 easily shows. Therefore { t, L 1, t} is a positive integer solution to the 1 diophantine equation y = z, with (,y)=1. Note that ( t1, t ) = 1 easily implies ( t, L ) = 1 = ( t, t ) in the above equation. Aordingly by Result we must have either κ, t = λ t = κ κλλ ; or alternatively t1= κ λ, t= κ κλ λ ; or positive integers κλ, suh that ( κλ, ) = 1 and κ λ 1(mod 4). (iv) Sine ρ must be a square, by (5) we know that ρ = n(m n) = square; and sine m-n is odd, we must have n 0(mod 4); whih means that n must be even. Additionally, by Proposition 1(ii), we must have m = t1 t and n = tt 1, with ( t, t ) = 1t t 1(mod), and t 1 > t. Altogether, n(m-n)=square L 1 1 ] L 4 tt( t t tt) = tti [( t t) = ; whih implies that, by Result 7, the positive integer ( t1 t) is a divisor of L; and so, L= ( t t ) i L1; going bak, 1 18

19 1 1 ] = 1 ) L1 1 = L1 t t i[( t t ) 4( t t i ; t t, and sine ( t1, t ) = 1, Result 4 (applied with n=) tells us that t 1 = κ and t = λ for some positive integers κ, λ suh that ( κ, λ ) = 1. But, by virtue of, t1 t 1(mod ), we must also have κ λ 1(mod ). (v) Here, we only have to observe that sine is a square, we must have, by Proposition 1(ii), α= m n, β= mn, = m n, m = t t, n = t, t,( t, t ) = 1, t t 1(mod ), and ρ tt 1 i t1 t and n (m-n)= tt 1 t1 t) β = n(m n) = 4 ( ) ( = ρ. And sine ( t1, t ) = 1, it is then lear that, ρ β = square ρ= square ( t1 = κ and t1 = λ ), for some positive integers κλ,, with ( κ, λ ) = 1andκλ 1(mod ). (vi) From formulas (5), we must have ρ = n(m n) = square; whih as we have seen previously in other parts of this Theorem s proof, easily implies that sine m-n is an odd integer; n must be even. But then β being a square implies (by Proposition 1(i)) that, m = t1, n = 1 t, t1 1(mod ),( t1, t) = 1,andt1 > t. We have, n(m-n)= L, for some L Z ; thus, ) 1 = ( t i ( t t ) L ; whih implies (by Result 7) that t is a divisor of L; and so L = ( t ) i L1, for some positive integer 1. L Going bak, ) 1 = ) L1 L1 = 1 ( t i( t t ) (t i t t, whih shows that { L 1, t, t1,} is a positive integer solution to the diophantine equation y = z, with (,y)=1 (note that ( t, t ) = 1 ( L1, t ) = 1) Hene, aording to Result 1 we must have, 1 λ t 1 t = κ, = κλ, for some positive integers κand λ, with κ odd and ( κλ, ) = 1. This onludes the proof.. 1) For Family 1. F Numerial Eamples 19

20 Eample 1: Take t t1 ) ρ α ) κ= 1, λ= 1; then = 1, =, m = 8, n = 1, α= 65, β= (4 = 16, = 63, = (1 = 144 Eample : Take t t1 ) ρα ) κ= 3, λ= ; then = 1, = 1,m = 88,n = 1, α= 8945, β= (4 = 576, = 8943, = (408 = ) For Family F Eample 3: Take t1 t ) ρβ ) κ= 1, λ=, then = 7, = 4, m = 49, n = 3, α= 345, β= (58 = 3364, = 1377, = (144 = 0736 Eample 4: Take 1 ρβ κ= 5, λ= 1; thent = 3, t = 10,m = 59,n = 00, α= , β= (460) = 11600, y = 39841, = (540) = ) For Family F 3 Eample 5: Take t1 t ) ρ ) κ= 1, λ= ; then = 5, = 7, m = 50, n = 49, α= 4901, β= (70 = 4900, = 99, = (10 = 100 Eample 6: Take t1 t ) ρ ) κ=, λ= 1; then = 5, = 1, m = 50, n = 1, α= 501, = 499, β= (10 = 100, = (70 = 4900 Eample 7: Take κ=, λ= 3 (use the alternative formula for t ); then t1 t ) ρ ) = 13, = 17, m = 338, n = 89, α= , β= (44 = , = 3073, = (18 = ) For Family F4= F5 0

21 Eample 9: Take 1 κ= 1, λ= ; thent = 1, t = 4, m = 17, n = 8, α= 353, β= 7, = (15) = 5, ρ= (1) = 144, ρ = (0) = 400 β Eample 10: Take 1 κ=, λ= 3; then t = 4, t = 9, m = 97, n = 7, α= 14593, β= 13968, = (65) = 45, ρ= (60) = 3600, ρ = (156) = 4336 β 5) For Family F 6 Eample 11: Take 1 κ= 1, λ= 1; thent = 3, t =, m = 9, n = 8, α= 97, β= (1) = 144, = 17, ρ= (4) = 16 Eample 1: Take 1 κ= 1, λ= ; thent = 9, t = 4, m = 81, n = 3, α= 7585, β= (7) = 5384, = 5537, ρ= (56) =

22 8. The diophantine equation y = z and losing remarks. A. A brief desription / sketh of the derivation of the general solution of the equation y = z with (,y) = 1 Below we outline a method for obtaining all the positive integer solutions to this equation, via two-parameter formulas. To this end, let {a,b,} be suh a solution. We must then have, a b = (a,b) = 1;anda,b, Z Obviously, beause of the ondition (a,b)=1; either a and b are both odd or otherwise they must have different parities. But 0or modulo 4 (for even or odd respetively). If a and b had different parity then, a b 1(mod 4). This learly shows that only the first possibility an our, namely a and b being both odd; and so being odd as well. Net, observe that the ondition (a,b)=1 easily implies (b,)=1=(a,); from whih it easily follows that any two of the three integers a,b,, must be distint; with a single eeption, the ase a=b==1. The above equation is the equivalent to a b ( ) ( ), = whih shows that the point ( a, b ) on the twodimensional XY plane lies on the irle of enter the origin (0,0) and radius : The irle with equation X Y =. The point ( a, b ) is a rational point (i.e. a point both of whose oordinates are rational numbers) on the irle. Assume that at least one of the positive integers a, b, is greater than 1. Whih means (aording to the above eplanation) that all three of them are pairwise distint and greater than 1. Then the point (1,1) lies on the same irle and it is distint from ( a, b ). Consider the straight line on the XY plane that passes through these two points. It must have

23 slope b 1 s = ; a 1 with s 0. Sine the point ( a, b ) lies on the above irle and in the first quadrant of the XY plane; it follows that s<0. The slope s is a negative rational number; so that s an be represented as K s = λ, where K and λ are positive integers, with (K, λ )=1. We have, b K 1 s = =. λ a 1 If we solve for b in terms of a, and K λ and substitute in the irle equation a b ( ) ( ) = ; after a ertain amount of algebra we arrive at, K a K K a K K ( ) 1 ( ) ( ) ( 1)( ) ( ) ( ) 1 = 0. λ i λ i λ λ λ The last equation shows that the number a is one of the two roots of the quadrati equation (in the variable t), K K K K K ( ) 1 it 1 i t ( ) ( ) 1 = 0. λ λ λ λ λ The other root is the number 1, as it an be seen by inspetion. But the sum of the two roots of a quadrati equation α t β t = 0, must equal β. Applying this fat to the above quadrati α equation whose roots are 1 and a ; and solving for a we obtain, a K Kλ λ = K λ Moreover, going bak to the equation for the slope s and solving for b we also obtain, b = K Kλλ K λ 3

24 At this point, the following simple result from number theory may be used: If a1, a, a3, a 4, are four positive integers suh that a a a = and ( a1, a ) = 1; then a3= δ a1 and a4=δ a; where a δ= ( a3, a4) (The greatest ommon divisor of 3 a and a 4). This typially an be assigned as a simple eerise in elementary number theory, whih an be easily proved with the aid of Result 3(i). In our situation we have a K = Kλ λ K λ and b = K Kλλ K λ Note that in fat, sine a,b,, and K λ are all positive; both integers K Kλ λ and λλ must also be positive. Also reall that (K, λ )=1 K K Now, by Result 3(iv), if δ= ( K K λ λ, K λ ), then δ = 1 or, respetively for K λ 1(mod ); or K λ 1(mod ). But if δ =, then aording to the above fat both integers a and would have to be multiples of, whih obviously annot happen, sine a and are both odd. The same argument applies to the seond fration; whih leads to the onlution that, a = K K λ λ,b= K K λ λ, = K λ ; and with (K, λ ) = 1 and K λ 1(mod ). Conversely, if a,b, satisfy these paremetri formulas, then a routine alulation shows that a b =. Even though some dubliation of solutions may our; to ensure that for any hoie of positive integers K and λ, the above formulas produe positive a and b ( is always positive), we may take a = K K λ λ,b= K K λ λ, = K λ. In onlusion, the entire family of positive integer solutions to the diophantine equation y = z an be desribed by, (,y) = 1 4

25 = K λ,y= K λ,z =, K λ K λ K λ where K, λ an be any positive integers; suh that (K, λ )=1, K λ 1(mod). Note that the solution {1,1,1} is the only one whih annot be obtained from the above formulas. B. Historial Remarks Aording to L.E. Dikson s book (see[1]), historially, the diophantine equation y = z has been investigated as a result of the problem of finding three integers whose squares are in arithmeti progression. A number of investigators are mentioned. Various researhers used a number of different tehniques and methods to takle this problem; most of them ahieved partial solutions or families of solutions, but not the general solution (i.e. entire family of solutions). Below we mention five researhers who worked on this problem, as ited in Dikson s book. Another twenty or so names of individuals are mentioned as ontributors to the understanding and solving of the above diophantine equation. Diophantus (ira 150AD-50AD) is reported to have disovered three partiular or speial numbers whose squares are in arithmeti progression. In the thirteenth entury, Jordanus Nemorarius found a partiular family of solutions: = b, y = b b, z = b b, where b an be any integer, while is an even integer. P. Fermat, in the 17 th entury, disovered the following family of solutions; =, y = 4rs, z = rs. r s r s r s 5

26 Then there is the 1918 paper of A.E. Jones, in whih he disovered a family of integer triples, suh that in eah triple, the squares of the three numbers are in arithmeti progression; and the sum of any two of the three numbers in the triple is a square as well. Finally, A. Desboves in the late 19 th entury, gave all the solutions to the more general diophantine equation, y = ( a b ) z (in our ase a=b=1) Referenes [1] Dikson, L.E., History of Theory of Numbers, Vol. II, AMS Chelsea Publishing, Providene, Rhode Island, 199 ISBN: ) ; 803 p.p. (unaltered tetual reprint of original book, first published by Carnegie Institute of Washington in 1919, 190, and 193) (a) For the diophantine equation (b) For the diophantine equation y = z, pages; 40, 41, 46 y = z, pages; 47, 48; (.) For Pythagorean triangles, pages ; for more general information on rational right triangles, pages [] Rosen, Kenneth H., Elementary Number Theory and its Appliations, Third Edition, 1993, Addison-Wesley Publishing Company (there is now a fourth edition as well), 544 p.p ISBN: (a) For Pythagorean triples, pages (b) For Result 3 (i), page 91, Lemma.3 () For Result 4, page 101, eerise 57 (d) For Result 7, page 100, eerise 35 6

27 [3] Sierpinski, W., Elementary Theory of Numbers, originals edition, Warsaw, Poland, 1964, 480 p.p. (no ISBN numbers). More reent version (1988) published by Elsevier Publishing, and distributed by North-Holland. North-Holland Mathematial Library 3, Amsterdam (1988) This book is available by various libraries, but it is only printed upon demand. Speifially, UMI Books on Demand, from: Pro Quest Company, 300 North Zeep Road, Ann Arbor, Mihigan, USA; ISBN: (a) For Result 3(i): pages 14,15 (b) For Result 4: pages 17,18 () For Result 7: page 15 [4] K. Zelator, The diophantine equation ky = z and the integral triangles with a osine value of 1, Mathematis and Computer Eduation, Fall 006 n 7