Math 319 Problem Set #2 Solution 14 February 2002

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1 Math 39 Problem Set # Solution 4 February 00. (.3, problem 8) Let n be a positive integer, and let r be the integer obtained by removing the last digit from n and then subtracting two times the digit ust removed. (See the hint in NZM for a nice way to formalize this operation.) Prove that 7 n if and only if 7 r. Proof: Let u be the units digit of n, and write n = 0m + u for some integer m. Then r = m u. Observe that so that 7 (n 3r). n 3r = 7m + 7u Now suppose 7 n. Then since 7 (n 3r), it follows by Theorem.(3) that 7 3r. Since (7, 3) = and 7 3r, we get from Theorem.0 that 7 r. For the converse, suppose 7 r. Then since 7 (n 3r) also, by Theorem.(3) we get 7 n.. (Based on.3, problem 6) Find a positive integer n such that n/ is a square, n/3 is a cube, and n/5 is a fifth power. Have you found the least such positive integer? Solution: Let n be our integer. Clearly n must be divisible by, 3, and 5, so to get started, let s write n = a 3 b 5 c. The given conditions tell us that a must be odd and a multiple of both 3 and 5; b must be of the form 3k + and must be a multiple of both and 5; and c must be of the form 5k + and must be a multiple of both and 3. That is, the number a must satisfy a (mod ), a 0 (mod 3), a 0 (mod 5). Any positive solution to this system must be divisible by [3, 5] = 5, and the least such number is a = 5. The number b must satisfy b 0 (mod ), b (mod 3), b 0 (mod 5).

2 Again, any positive solution to this system must be divisible by [, 5] = 0, and the number 0 happens to be a solution. Finally, the number c must satisfy c 0 (mod ), c 0 (mod 3), c (mod 5). That is, c must be a multiple of [, 3] = 6, and it happens that c = 6 satisfies all three congruences. Taking a = 5, b = 0, and c = 6, we get n = = 30, 33, 088, 000, 000. This is the smallest such number because the three eponents we chose were all the least positive integers satisfying their respective criteria. 3. (Based on.3, problem 7) Let P be the set of pairs of twin primes greater than. That is, let P = {(3, 5), (5, 7), (, 3), (7, 9),...}. Let N be the set of positive integers n > 3 such that n has eactly four positive divisors. Prove that there is a one-to-one correspondence between P and N. Proof: For each pair (p, p + ) P, let Φ(p, p + ) = p +. Then if n = Φ(p, p + ), we have n = (p + ) = ((p + ) )((p + ) + ) = p(p + ), the product of two distinct primes. Thus there are eactly four positive divisors of n, namely,, p, p +, and p(p + ). To see that these divisors are all distinct, we need only point out that p > and > 0. Together these imply that < p < p + < p(p + ). Furthermore, since p 3, we have p + > 3. Therefore, n N. This shows that the range of Φ is a subset of N.

3 The map Φ is clearly one-to-one, because if Φ(p, p+) = Φ(q, q+), then p+ = q+, from which it follows that (p, p + ) = (q, q + ). The map Φ is also onto. Let n N. Write n = (n )(n + ). Since n > 3, we have n >. Multiply both sides by the positive integer n + to get This establishes that n > n +. < n < n + < n, so that these four integers are all distinct and positive. They are also all divisors of n, and since n N, they must be the only four positive divisors of n. It follows that n must be prime, for otherwise it would have a prime divisor p between and n, and then p would be a fifth positive divisor of n. Similarly, n + can have no proper divisors other than n, because any such divisor would be a fifth positive divisor of n. If (n ) (n + ) for positive n, we must have (n ) n +, from which it follows that n 3. But we know n > 3, so n cannot be a factor of n +, and thus n + is prime. We have shown that if n N, then n and n + are both primes, and since n > 3, we have n 3, so the pair (n, n + ) is an element of P. Finally, we have Φ(n, n + ) = n, and since n was an arbitrary element of N, we have shown that Φ is onto. Thus the map Φ is the desired one-to-one correspondence between P and N. 4. (Part of.3, problem 9.) Let a and b be positive integers such that (a, b) = and ab is a perfect square. Prove that a and b are perfect squares. Proof: By the Fundamental Theorem of Arithmetic, we may write a = p e p e p e k k b = p f p f p f k k

4 where p, p,... is the sequence of primes, and all eponents are zero once we get far enough out in the sequence. Since ab is a perfect square, we know that e i + f i is even for every i. Since (a, b) =, we know that, for each i, only one of e i and f i is non-zero. Now for each i, we have that e i + f i is even, and either e i = 0 (in which case f i must be even) or f i = 0 (in which case e i must be even). In either case, both e i and f i are even (the number 0 is even, despite the deeply-held beliefs of certain calculus students), and it follows that a and b are both perfect squares. 5. (.3, problem 3 also read problem 30 and the remarks following problem 3) Prove that no polynomial f() of degree greater than (or equal to) with integral coefficients can represent a prime for every positive integer. Proof: Write f() = a 0 + a + a + + a n n. Suppose f() is prime for every positive integer. Let p = f(). Then p is prime. Now let k be a positive integer, and consider f( + kp) = a 0 + a ( + kp) + a ( + kp) + + a n ( + kp) n. Note that for each, ( + kp) = + = + pn where N is the integer given by kp + (kp) + (kp) + (kp) N = k + k p + + k p + k p. Thus we may write f( + kp) = ( + pn 0 )a 0 + ( + pn )a + + ( + pn n )a n = a 0 + a + + a n + a 0 (pn 0 ) + a (pn ) + a (pn ) + + a n (pn n ) = f() + p(n 0 + N + + N n ).

5 In fact, since f() = p, we have so that p f( + kp) for all integers k 0. f( + kp) = p( + N 0 + N + + N n ) Now by hypothesis, f( + kp) is prime for every k 0, so it follows that f( + kp) = p for every k 0. Consider g() = f() p. Then g() is a polynomial and g( + kp) = 0 for every k 0. That is, g has infinitely many roots. So g must be identically 0. Thus f must be identically equal to p. But this contradicts the assumption that the degree of f is greater than or equal to. 6. (Part of.3, problem 53) Let π() denote the number of primes not eceeding. Show that p p = π() + du u where the sum is taken over all primes p less than or equal to. Proof: First we note that for a positive integer k >, { if k is prime π(k) π(k ) = 0 if k is composite. Net we observe that is constant in any interval [n, n) where n is an integer. Thus we get We write n n du = π(n ) u [ = π(n ) n n du u n n ]. 3 4 du = du + du + + u u 3 u [ = π() [ + π(3) 3] 3 + 4] [ +π( ) ] + π( ) du + du u u [ ],

6 where the factor π( ) comes out of the final integral because is constant on [, ). We collect multiples of each reciprocal /n to get u du = π() + 3 (π(3) π()) + (π(4) π(3)) (π( ) π( )) π( ). Net we use the fact that π() = and π(n) π(n ) is if n is prime and 0 if n is composite to write du = u p p π( ) where the sum is taken only over the primes. Finally, we note that π( ) = π() and move the rightmost term to the left side of the equation to conclude that u du + π() = p p as required.

Homework #2 solutions Due: June 15, 2012

Homework #2 solutions Due: June 15, 2012 All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is