PRACTICE PROBLEMS: SET 1

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1 PRACTICE PROBLEMS: SET MATH 437/537: PROF. DRAGOS GHIOCA. Problems Problem. Let a, b N. Show that if gcd(a, b) = lcm[a, b], then a = b. Problem. Let n, k N with n. Prove that (n ) (n k ) if and only if (n ) k. Problem 3. Let a, b, n N \ {} with a b. Prove that (a n b n ) (a n + b n ). Problem 4. Let a, b Z such that gcd(a, b) =. Prove that gcd(a + b, a ab + b ) {, 3}. Problem 5. Let x, y, n N \ {} such that n + = xy. Prove that for each a N, we have a (x ) if and only if a (y ). Problem 6. 4n + 3. Prove that there are infinitely many prime numbers of the form Problem 7. Let n N be a composite number larger than 4. Prove that n (n )!. Problem 8. Let f Z[x] be a nonconstant polynomial. Prove that there exists n N such that f(n) is composite. Problem 9. Let n N \ {}. Prove that n is composite. Problem 0. Let n N. If n + is a prime number, then n is a power of. Problem. Show that 4 is the largest integer divisible by all integers less than its square root. Problem. How many intergers are between 408 and 6666 which are divisible by, but they are not divisible by neither 3 nor 5? Problem 3. Prove that the product of any four consecutive integers is divisible by 4. Problem4. (a) Let x Z \ {}, and let n N. Prove that (x ) (x n ).

2 MATH 437/537: PROF. DRAGOS GHIOCA (b) Let a, b N. Prove that gcd ( a, b ) = gcd(a,b). (c) Let a, b N such that b 3. Prove that ( b ) ( a + ). Problem 5. Let n N. Prove that gcd (n! +, (n + )! + ) =. Problem 6. Let n N \ {}. Show that composite. ( ) n +n n! if and only if n + is Problem 7. Let n N \ {}. Prove that n 4 + n + is composite. Problem 8. Let n N \ {}. Prove that n i= i is not a positive integer. Problem 9. Prove that an odd integer n > is a prime if and only if it is not expressible as a sum of three or more consecutive positive integers. Problem 0. Let a, b Z such that a b. Prove that (a b ) (a (a b b) ). Problem. Prove that there exists a subset A of {,, 3,..., 6 } with at most 80 elements such that, 6 A and each element of A \ {} is the sum of two elements of A (not necessarily distinct). Problem. Prove that the function f : N {0} N {0} defined by is surjective. f(n) = [n 3] [n ] Problem 3. Find all positive integers n such that n! = n + 8. Problem 4. Let a be an integer larger than. We construct the sequence: Find all a such that x 03 N. x 0 = a and x n+ = ax n, for all n 0. Problem 5. If p and q are prime numbers satisfying { p} = { q}, then prove that p = q. Problem 6. Find all positive integers m and n such that 8 m +9 n + is a perfect square. Problem 7. Let m N. Show that there exists n N such that n m = [ 03 n ] + [ 3 n ] +.

3 PRACTICE PROBLEMS: SET 3. Solutions. Problem. Let d = gcd(a, b) and let M = lcm[a, b]. Since d, a, b N then d min{a, b}. Similarly, because a M and b M, then max{a, b} M. So, because d = M we get min{a, b} = max{a, b}; therefore a = b. Problem. We have So, (n ) (n k ) if and only if k n k = (n ) n i. i=0 k (n ) n i. Claim. For each i N we have (n ) (n i ). Proof of Claim. We prove the above Claim by induction on i. The base case i = is obvious. So, assume now that (n ) (n i ) and we will prove that i=0 (n ) ( n i+ ). Indeed, n i+ = n i (n ) + (n i ) is a multiple of n (we also use the inductive hypothesis here). This concludes the proof of the Claim. Using the above Claim we get that k k n i k = (n i ) i=0 is a multiple of n always. So, i= k (n ) if and only if (n ) k, as desired. In the above reasoning we also used the fact that if a x, then a x + y if and only if a y. Problem 3. We prove the result by contradiction. Without loss of generality, we may assume that a > b, and let d = gcd(a, b). Then a = d a and b = d b, where gcd(a, b ) =. We have that i=0 n i (a n b n ) (a n + b n ) if and only if a n b n d n an + b n d n. Note that d n a n and d n b n which yields that d n a n b n and d n a n + b n. So, (a n b n ) (a n + b n ) if and only if (a n b n ) (a n + b n ). Clearly (a n b n ) (a n b n ) which yields that (a n b n ) b n.

4 4 MATH 437/537: PROF. DRAGOS GHIOCA Claim. gcd(a n b n, b ) =. Proof of Claim. Let p be any prime number dividing b. It suffices to prove that p (a n b n ). Indeed, if p (a n b n ) and also p b n (because p b ) we obtain that p a n. Since p is a prime number we conclude that p a. But since p b and gcd(a, b ) =, we obtain a contradiction; this concludes the proof of our Claim. Using that gcd(a n b n, b ) = we obtain that gcd(a n b n, b n ) =. But we know that (a n b n ) b n, which yields that (a n b n ). Since a > b we know that a > b ; also we know that n. So, a n b n = (a b ) ( a n + a n b + + a b n + b n ) a n + a n b + + a b n + b n + = 3. Note that b and a was used in establishing the above inequality. Because a n b n >, then (a n b n ), which yields a conatradiction with the above assumption. Therefore (a n b n ) (a n + b n ) as desired. Problem 4. We have Since (a + b) (a + b) we get that a ab + b = (a + b) 3ab. gcd(a + b, a ab + b ) = gcd(a + b, 3ab). Note that in establising the above equality we used the fact that for any three integers x, y, z, if x z, then gcd(x, y) = gcd(x, y + z). Because of this observation, we also note that = gcd(a, b) = gcd(a + b, b) = gcd(a, a + b). Therefore gcd(a + b, ab) =. So, the only possible common divisors of a + b and of 3ab are and 3. Problem 5. We start by proving the following. Claim. max{x, y} < n. Proof of Claim. Without loss of generality, we assume x y. So, it suffices to prove that y < n. We know that x and also that n which yields that n >. So, using that n+ = n + n > n + = xy y, we conclude that y < n, as desired. Using the above Claim we obtain that if a (x ) then a < n and thus a n. Similarly, if a (y ) then a n. Note that we are also using that x > and y > which guarantees that x and y are positive integers which allows us to employ the inequalities a x, respectively a y. Now, assume a x. As noted above, a < n, and thus a n (xy ).

5 PRACTICE PROBLEMS: SET 5 We write xy = (x )y + (y ). So, because a (x ) and a (xy ), we conclude that a y. The exact same argument with the roles of x and y reversed yields that if a (y ), then a (x ). Problem 6. We proceed by contradiction, and assume there are finitely many prime numbers of the form 4m + 3: Let 3 = p, 7 = p, = p 3,..., p k. N = 4p p 3 p k + 3. Note that the above product is not empty since there are indeed primes of the form 4m + 3 larger than 3. Claim. For each i =,..., k, we have p i N. Proof of Claim. The proof for i = is somewhat different than the proof for i >. Indeed, if 3 were a divisor of N, then 3 4p p 3 p k. But 3 4 and also 3 p i for i since each p i is a different prime number and 3 = p < p i for i. Hence p N. Now, assume p i N for some i. In this case we know that p i 4p p 3 p k and thus p i must divide 3 as well. However, for i, we know that p i > 3 (as explained above); hence p i N for i, which concludes the proof of Claim. By construction, N is an odd integer larger than. Therefore there exists at least one prime number dividing N; furthermore, all primes dividing N are odd. Claim. There exists a prime p of the form 4m + 3 which divides N. Proof of Claim. We already know that all prime divisors of N are odd. Assume none of them is of the form 4m + 3; therefore they are all of the form 4m +. Now, the product of numbers of the form 4m + is also of the same form. This last statement suffices to be proven for a product of two factors (and then use induction on the number of factors in an arbitrary product of numbers of the form 4m + 3). Now, indeed (4m + )(4m + ) = 4(4m m + m + m ) +, which shows that any product of numbers of the form 4m + is of the same form. Therefore, it cannot be that all prime factors of N are of the form 4m + since then N would be of the same form (and we know it isn t). Hence there exists a prime number p of the form 4m + 3 which divides N. This concludes the proof of Claim. So, N is divisible by a prime p of the form 4m + 3 (by Claim ), and p is none of the prime numbers p i from our list (by Claim ). This shows that there are indeed infinitely many prime numbers of the form 4m + 3. Problem 7. Since n is composite, then n = ab for some positive integers a and b larger than. Case. a b.

6 6 MATH 437/537: PROF. DRAGOS GHIOCA We know that a > and b > ; so, from n = ab, we also get that a < n and b < n. Now, by the assumption in Case, we know that a and b are distinct integers between and n. Therefore n = ab (n )!, as desired. Case. a = b. In this case, n = a and since n > 4 we conclude that a >. Hence a < a = n. Therefore a and a are two distinct integers between 3 and n (note that a = n < n for n > ). So, a (a) (n )!. In particular, n = a (n )! as desired. Problem 8. Assume f(m) is a prime number p, for some m N. Claim. For each k Z, we have that p f(m + pk). Proof of Claim. We have that f(x) = d i=0 c ix i, for some c i Z, where d = deg(f). Since f(m) = p, it suffices to prove that p (f(m + pk) f(m)). In order to prove this last statement we prove that p ( (m + pk) i m i) for each i 0. Indeed the above statement is obvious for i = 0 and i =, while for i we have (m+pk) i m i = pk ((m + pk) i + (m + pk) i m + + (m + pk)m i + m i ). Thus the above divisibility holds for all i 0, and so indeed p f(m + pk) f(m). This concludes the proof of Claim. Now, for each k N if f(m + pk) were to be a prime number, then it would actually have to be equal to p (since it is divisible by p). However, the equation f(x) p = 0 cannot have more than d = deg(f) roots since d, i.e., f is nonconstant. Therefore, not all numbers f(m + pk) (and more generally not all numbers f(n)) can be prime numbers. We actually proved that f(x) takes infinitely many values which are composite. Problem 9. We have n = n 4 + 4n + 4 4n = (n + ) (n) = (n n + )(n + n + ). Therefore all we need to prove is that the numbers n n + and n + n + are larger than. In particular, it suffices to prove that This fact is true because since n >. n n + >. n n + = (n ) + >, Problem 0. Clearly, if n =, then n is a power of more precisely, n = 0. So, from now on we assume n. For any odd integer k >, and any numbers a and b we have a k + b k = (a + b) (a k a k b + a k 3 b a k 4 b 3 + ab k + b k ).

7 PRACTICE PROBLEMS: SET 7 Assume now that n is divisible by some odd prime number p. We let a = n p and thus n + = a p + = (a + ) (a p a p + a p 3 + a + ). Since n + > a + (because n > n p ), we conclude that n + is divisible by a +, which is a number less than n + and also larger than. Therefore n + is not a prime number. In conclusion, n has to be a power of (since it s not divisible by any odd prime number) in order for n + be a prime number. Problem. We have to prove that if n 5, then n cannot be divisible by all numbers less than its square root. We let m be the unique positive integer satisfying m < n (m + ). Assume n is divisible by all integers between and m; we will prove that this contradicts the fact that n 5. Case. n = 5. In this case, m = 4, and clearly 5 is not divisible by 4. Case. 6 n 36. In this case m = 5. So, n would have to be divisible by,, 3, 4 and 5; thus 60 n, which is impossible. Case n 49. In this case m = 6, and so, n would have to be divisible by,, 3, 4, 5 and 6; thus 60 n, which is impossible. Case 4. n > 49. Assume n is indeed divisible by all integers between and m; note than m 7 since n > 49. The following easy claim will be useful. Claim. If k 3 is an odd integer, then k, (k ) and (k ) are all relatively prime. Proof of Claim. Clearly, gcd(k, k ) = gcd(k, k ) =. Also the only possible common divisor of k and k must be also a divisor of, but it cannot be since k is odd; therefore gcd(k, k ) =, which concludes the proof of our Claim. Now, if m is odd, then m(m )(m ) n because of the above Claim. If m is even, then (m ) is odd, and therefore (m )(m )(m 3) n (again by the above Claim ). In either case we must have (m )(m )(m 3) n. The following result shows that the above inequality contradicts the fact that m 7 and that n (m + ). Claim. Let x R. If x 7, then (x )(x )(x 3) > (x + ). Proof of Claim. We will show that f(x) = (x )(x )(x 3) (x + ) is positive for x 7. So, and since x 7, we have as desired. f(x) = x 3 7x + 9x 7, f(x) = x (x 7) + (9x 7) 56 > 0,

8 8 MATH 437/537: PROF. DRAGOS GHIOCA Therefore we proved that also in Case 4 we obtain a contradiction once we assume that n is divisible by all integers between and m. This concludes the proof of Problem. Problem. For each positive integer k we denote by N k the number of integers between 408 and 6666 which are divisible by k. Then the number sought in this question equals: N N 6 N 0 + N 30. Indeed, we have to subtract N 6 +N 0 since we have to exclude from our computation the numbers divisible by 3 or 5 (and also divisible by ). Finally, we need to add N 30 since in N 6 + N 0 we counted twice the numbers divisible by both 6 and 0. We compute N = = 630 (the extra from the formula comes from the fact that both 408 and 6666 are even numbers). We compute N 6 = = 877. The reasoning behind this computation is that both 6666 and 40 are divisible by 6 (this explains the + ) and also we used 40 (and not 408) since this is the smallest integer at least equal to 408 which is divisible by 6. We compute with the same reasoning: N 0 = In conclusion, the answer is N 30 = = 56 and + = = 403. Problem 3. Let n Z; we have to show that 4 n(n + )(n + )(n + 3). Since 4 = 3 8 and gcd(3, 8) =, it suffices to prove that and also 3 n(n + )(n + )(n + 3) 8 n(n + )(n + )(n + 3). Among any three consecutive integers, there exists one integer divisible by 3; therefore, 3 n(n + )(n + ) n(n + )(n + )(n + 3). Similarly, among any four consecutive integers there exists one integer of the form 4m and another integer of the form 4m + ; thus one of the four consecutive integers is divisible by 4 and another one is divisible by. Their product is divisible by 8, which proves that This concludes our proof. Problem 4. 8 n(n + )(n + )(n + 3). (a) We have x n = (x )(x n + + x + ), which yields that (x ) (x n ).

9 PRACTICE PROBLEMS: SET 9 (b) We will prove that for each N N \ {}, gcd(n a, N b ) = N gcd(a,b). We let d = gcd(a, b). So, a = da and b = db for some positive integers a and b. Applying part (a) with x = N d and n = a (respectively n = b ) we obtain that (N d ) (N a ) and (N d ) (N b ). Hence (N d ) M = gcd(n a, N b ). We prove next that M N d. Since d = gcd(a, b), then d is a linear combination of a and b. Without loss of generality we may assume there exists u, v N such that d = au bv. Applying again part (a) above with x = N a and n = u (respectively n = v) we get that M (N au ) and M (N bv ). Using that au = d + bv, we compute N au = N d (N bv ) + (N d ). Since M gcd(n au, N bv ) we obtain that M N d. Because (N d ) M and M N d we conclude that M = N d and thus, part (b) holds. (c) Assume the contrary, i.e., ( b ) ( a +). Because a = ( a )( a +) we conclude that ( b ) ( a ). By part (b) above, we conclude that b a. Since b >, we obtain that d = gcd(b, a) (otherwise gcd(a, b) = and then b which is a contradiction with the fact that b 3). But then ( d ) = gcd( a, b ), again by part (b) above. Since we assumed that ( b ) ( a + ), we obtain that also ( d ) ( a + ). From the two divisibilities: ( d ) ( a ) and ( d ) ( a + ) we obtain that ( d ). However d which yields a contradiction with the last divisibility. Therefore ( b ) ( a + ). Problem 5. Let p be a prime number dividing both n! + and (n + )! +. Then p divides their difference: (n + )! n! = n! n. So, p n! or p n; either way, p n!. But since p (n! + ), we would get that p, which is impossible. So, gcd (n! +, (n + )! + ) =. Problem 6. We let p = n + and we prove that if p is a prime, then p(p ) (p )!,

10 0 MATH 437/537: PROF. DRAGOS GHIOCA and that if p is composite, then p(p ) (p )!. For the first implication above, we note that p 3 since n. Because we assumed that p is a prime number, then p is an odd prime number. Therefore p is even, and thus if p(p ) (p )!, we would have that also p (p )! (since p N). Because p is a prime number, then none of the positive integers less than p are divisible by p. So, indeed, if p is prime then p(p ) (p )!. Now, assume p is composite. We can easily check that p(p ) (p )! for p = 4 (we get that 6 6). So, assume from now on that p 6 (note that p is composite and thus it cannot be equal to 5). In this case we will prove an even stronger statement: p(p ) (p )!. For the last divisibility it suffices to prove that p (p )!. We know that p is composite; therefore there exist integers a and b larger than such that p = ab. Without loss of generality, we assume a b. We split our proof now in two cases. Case. a < b Because a, we have b = p a p p, where in the last inequality we used the fact that p 6 and thus p >. Therefore a and b are two distinct integers between and (p ) and so, they appear in the product from (p )!.Hence p = ab (p )!, as desired. Case. a = b. In this case a = p; furthermore p is a perfect square in this case and because we already assumed that p 6, we conclude that p 9. Claim. If x R satisfies x 9, then x x. Proof of Claim. The function f(x) = x x is strictly increasing for x 9 since Therefore f (x) = x 6 > 0. f(x) f(9) = 7 3 = > 0. So, using the above Claim we get that a and a are distinct integers between and p. Hence their product divides (p )!. So, p = a a (a) (p )!, as desired. In conclusion, also in Case we have that p (p )!; this concludes the proof of Problem 6.

11 PRACTICE PROBLEMS: SET Problem 7. We note that n 4 + n + = (n 4 + n + ) n = (n + ) n = (n n + )(n + n + ). Since n >, we have that n n + = n(n ) + >, and also n + n + >. Therefore n 4 + n + is a product of two factors both larger than, and hence it is composite. Problem 8. We let k be the unique positive integer satisfying k n < k+. Then there exists a unique fraction i in the given sum which contains the largest power of precisely k. Indeed, has this property, and clearly there is no k fraction i in the above sum which has a denominator divisible by k+ because then i k+ > n, which is a contradiction. But also, there is no other fraction i where the denominator i is divisible by k (besides the fraction ). Indeed, if k i, and i k (but k i as in the above sum), then i k > n, which again contradicts the fact that i n. Hence is the unique fraction from k the above sum which has the largest power of appearing in the denominators of all fractions added in our sum. Now we let N = lcm[,,..., n] and then bring all the fractions i to a common denominator. Then the amplifying factor for all fractions, with the exception of is even, since all fractions, with the exception of have a power of in the k k denominator which is inferior to k. Therefore, for each i {,..., n} \ { k }, we have that On the other hand, i = N i N, where N i is an even integer. k = N k N, where N is an odd integer, k since the largest power of in N is indeed k (note that N is the least common multiple of all the numbers between and n, and there exists no higher power of than k appearing as a divisor for these numbers). In conclusion, n M = N i is odd, and thus n i= i= i = n i= since M is odd, while N is an even number. N i N = M N / N, Problem 9. Since n >, we have to prove that n is composite if and only if n is a sum of three or more consecutive integers.

12 MATH 437/537: PROF. DRAGOS GHIOCA Assume first that n is a composite number. We know (from hypothesis) that it is odd, and thus n = ab for some odd integers a and b larger than. Without loss of generality, we may assume a b. In particular, b = k + with k. Then n = ab = a (k + ) = (a k) + (a k + ) + (a ) + a + (a + ) + + (a + k ) + (a + k). We note that a k > a b 0, and thus n is indeed the sum of k + 3 consecutive positive integers. Assume now that n is a sum of l 3 consecutive positive integers. Case. l is odd. Then l = k +, and we may write the consecutive positive integers as a k, a k +,..., a, a, a +,..., a + k, a + k. Then we obtain that n = a (k + ) (exactly as above). Also, since a k (and k because l = k + 3) we obtain that a. Therefore n is a product of two numbers (a and l) which are both larger than. So, n is composite. Case. l is even. Then l = k with k (since l 3). We write the consecutive positive numbers such as a k +, a k +,..., a, a +,..., a + k, a + k. Therefore n = (a + ) k. Once again, we know that both factors of n are larger than since a + > a and also k (as observed above). In conclusion, again we obtain that n is composite, which concludes our proof. Problem 0. Clearly, (a b b) ( b) = b (mod (a b )). Therefore, as desired. a (a b b) a b 0 (mod (a b )), Problem. The set A consists of all numbers: In the above list there are: 6 6 i 3 i for all i = 0,, i 6 i for all i = 0,,, i 8 i for all i = 0,,, i 4 i for all i = 0,,, 4 3 4, 3,, = 74 numbers.

13 PRACTICE PROBLEMS: SET 3 We also note that 6 A and also A. Finally, each number of A other than is a sum of two numbers contained in A (not necessarily distinct). Indeed, 6 = ( 6 3) + ( 3 ) ( 6 i 3 i = 6 (i+) 3 (i+)) ( + 6 (i+) 3 (i+)) for each i = 0,,, 30 3 i 6 i = 3 = ( 3 6) + ( 6 ) ( 3 (i+) 6 (i+)) for each i = 0,,, 5 ( 3 (i+) 6 (i+)) + 6 = ( 5 ) = ( 5 ) + 5 = ( 5 8) + ( 8 ) ( 5 i 8 i = 5 (i+) 8 (i+)) ( + 5 (i+) 8 (i+)) for all i = 0,,, 7 7 i 4 i = 8 = ( 7 ) = ( 7 ) + 7 = ( 7 4 ) + ( 4 ) ( 7 (i+) 4 (i+)) + ( 7 (i+) 4 (i+)) for each i = 0,,, 3 4 = ( 3 ) = ( 3 ) + 3 = = 3 + ; 3 = + ; = +. Problem. It suffices to show the following claims: (a) f(0) = 0 and f(n) as n. (b) f(n) f(n + ) f(n) +. First let s explain why (a)-(b) yield the desired conclusion. Condition (b) says that there are no jumps of more than one in {f(n)} n N {0}, while condition (a) yields that {f(n)} n N {0} is unbounded (starting from 0). Part (a) is simple because n 3 n and always because f(n) > n( 3 ) [n 3] > n 3 and [n ] n. Part (b) is harder. First we prove the left inequality (which is slightly easier). Indeed, it suffices to show that For this last inequality we notice that f(n) (n + ) 3 [(n + ) ]. (n + ) 3 [(n + ) ] [n 3] + [n ] + = (n 3 [n 3]) ([(n + ) ] [n ]) 3 + > 0.

14 4 MATH 437/537: PROF. DRAGOS GHIOCA In the last inequality we used the fact that [(n + ) ] [n ] < (n + ) [n ] = + (n [n ]) < + < 3, and therefore So, [(n + ) ] [n ]. Now, for the right inequality, it suffices to prove that (n + ) 3 [(n + ) ] < f(n) +. (n + ) 3 [(n + ) ] [n 3] + [n ] = (n 3 [n 3]) ([(n + ) ] [n ]) + 3 In the last inequality we used the fact that because [(n + ) ] [n ] < + 3 <. [(n + ) ] = [n + ] [[n ] + ] [n ] +. Problem 3. We prove that for all n 5 we always have n! > n + 8. Indeed, we note that for all n 5 we have n! n(n ) 3 = 6n 6n. Then we prove that for all n 5 we have This is equivalent to proving that Indeed, we have that for all n 5: 6n 6n > n n 6n 8 > 0. 5n 6n = n(5n 6) n 9 95 > 8. So, there exist no solutions to n! = n + 8 for n 5. Then for all n =,, 3, 4 we check whether there is a solution and we find that the only solution is n = 4 since 4! = 4 = Problem 4. We claim that x 03 N if and only if there exists an integer b larger than such that a = b 04. First of all, if a = b 04, then for each n = 0,..., 03 we have x n = b n. We prove this claim by induction on n: the case n = 0 is immediate. Now, assume x n has the above form (for some n < 03) and then (by the inductive hypothesis) x n+ = ax n = b 03 b n = b (n+), as desired. So, if b N, then x 03 N.

15 PRACTICE PROBLEMS: SET 5 Now, we assume x 03 N and then we prove there exists an integer b larger than such that a = b 04. We let b = a 04 and we ll show that b N (note that since a > then also b > ). Argued as above, we conclude that x 03 = b = b 04 = a b. Since x 03 N (and also a N) then b Q. The next claim finishes our proof. Claim.. If a and k are integers larger than and k a Q then there exists b N such that a = b k. Proof of Claim. So, we know there exist x, y N (they re positive since a > ) such that k a = x/y and so, ay k = x k. Therefore for each prime number p, the exponent of p in a (denoted exp p (a)) satisfies exp p (a) = k (exp p (x) exp p (y) ). Hence k exp p (a) for each prime number p; thus k a N, as desired. Problem 5. We write { p} = p [ p] and same for q, and therefore conclude that p q = [ q] + [ p] Z. We square and conclude that Thus pq = k N which yields that p + q pq Z. 4pq = k. Now, if p q, then one of them has to be odd; without loss of generality assume q p. Then the exponent of q in 4pq is exactly. However, this contradicts (using the Fundamental Theorem of Arithmetic) the fact that 4pq = k and so, the exponent of q (or for that mater of any prime number) in k has to be even. Problem 6. Assume x = 8 m + 9 n + for some x N. Assume also that m. Then since 8 m + 9 n + is even, and also perfect square, then x is even and so 4 x and therefore 8 m + 9 n + is divisible by 4. But because we assumed that m, then 4 8 m = 9 m m. Hence 4 9 n +. But n ( ) n 9 n + = (8 + ) n + = k = + 8l, k for some integer l. This contradicts the fact that 4 9 n +, and so, 8 m + 9 n + is not a perfect square if m. So, m must be equal to if 8 m + 9 n + = x for some positive integer x. Then k= n + = x yields (x 3 n )(x + 3 n ) = 9. Therefore, x 3 n = and x + 3 n = 9 and therefore the only solution is (m, n) = (, ).

16 6 MATH 437/537: PROF. DRAGOS GHIOCA Problem 7. Let n (3m) 03. Then [ 03 n] + [ 3 n] + n 03 n+ 3 n+ n < 03 n + 03 n + 03 n 3m + 3m + 3m = m. So, let [ 03 n] + [ 3 n] + A = {n N : n m }. Then A is a bounded set of integers (since its largest element is less than (3m) 03 ); also A is nonempty because A (note that 3 /m). So, let n m be the largest element of A. We claim that n m m = [ 03 n m ] + [ 3 n m ] +. Indeed, if we assume we have strict inequality (note that n m A) then we ll prove that also n m + A which will then contradict the fact that n m is the largest element of A. So, we assume that and then we get that m < [ ] [ ] 03 n m + 3 nm + m ([ 03 n m + ] + [ 3 n m + ] + ) m ([ ] [ ] ) 03 n m + 3 nm + But then (since it s an integer) n m > n m. m ([ 03 n m + ] + [ 3 n m ] + ) n m +, and thus n m + A which gives a contradiction. So, indeed n m m = [ 03 n m ] + [ 3 n m ] +.