Solutions to Practice Problems, Math 312 1 Prove that the fourth power of an odd integer is expressible in the form 16n + 1 for n 2 Z. Solution If m is odd, we can write m = 2k + 1 for some k 2 Z and so m4 = (2k + 1)4 = 16k4 + 32k3 + 24k2 + 8k + 1 and so we have k(3k + 1) m = 16 k4 + 2k3 + + 1: 2 k(3k+1) k It remains to show that 2 2 Z. If k is even, then 2 2 Z, while if k 3k+1 is odd, then 3k + 1 is even and so 2 2 Z. This completes the proof. 2 Define an by a0 = 1; a1 = 2; a2 = 4 and an+2 = an+1 + an + an−1; for n ≥ 1: n Show that an ≤ 2 for all n 2 N. Solution Note the typo - I meant to write a2 = 4. We use induction on n. The inequality is true for n = 0; 1 and 2. Suppose that it is true for all n ≤ k where k ≥ 2. Then k k−1 k−2 k−2 k+1 ak+1 = ak + ak−1 + ak−2 ≤ 2 + 2 + 2 = 7 · 2 < 2 : 3 If Fn is the nth Fibonacci number, prove that 2 n Fn+1Fn−1 − Fn = (−1) : Solution We use Binet's formula and find that the left-hand-side is αn+1 − βn+1 αn−1 − βn−1 αn − βn 2 p p − p 5 5 5 and hence is equal to 2(αβ)n − αn+1βn−1 − αn−1βn+1 : 5 Since αβ = −1, this is equal to 2 − αβ−1 − α−1β 2 + α2 + β2 (−1)n = (−1)n = (−1)n: 5 5 4 Let a and n be positive integers with a > 1. Prove that, if an + 1 is prime, then a is even and n is a power of 2. Solution If an + 1 is prime, then, since a > 1, it follows that an + 1 must be odd and so a must be even. If n > 1, then if n is not a power 1 2 of 2, we would have that n is divisible by some odd prime, say p. If n = p, we have an + 1 = ap + 1 = (a + 1) ap−1 − ap−2 + · · · − a + 1 and so an + 1 is not prime (each factor above is > 1). If n > p, an +1 = (an=p)p +1 = (an=p +1) a(n(p−1)=p − an(p−2)=p + · · · − an=p + 1 which again contradicts the fact that an + 1 is prime. 5 Show that if all three of p; p + 2 and p + 4 are prime, then p = 3. Solution We apply the Division algorithm to p and 3 and find that p has a remainder of either 0, 1 or 2, after division by 3. In the first case, we necessarily have p = 3. In the second, it follows that p + 2 is divisible by 3 and hence equal to 3, contradicting the fact that p is prime. In the third case, we have that p + 4 is divisible by and hence equal to 3, again a contradiction. 6 Use the Euclidean algorithm to compute (2059; 2581) and to express this quantity as a linear combination of 2059 and 2581. Solution We have 2581 = 1 · 2059 + 522 2059 = 3 · 522 + 493 522 = 1 · 493 + 29 493 = 17 · 29 + 0 and so (2059; 2581) = 29. We thus can write 29 = 522 − 493 = 522 − (2059 − 3 · 522) = 4 · 522 − 2059 and so 29 = 4(2581 − 2059) − 2059 = 4 · 2581 − 5 · 2059: 7 Show that every nonzero integer can be uniquely expressed as k k−1 ak3 + ak−13 + ··· + a13 + a0 where ai 2 {−1; 0; 1g and ak 6= 0. Solution This follows from the Division algorithm upon noting that if, after dividing n by 3, we obtain a remainder of 2, we may write n = 3k + 2 = 3(k + 1) − 1: 8 Prove that there are infinitely many primes of the shape 6n + 5. Solution We note that if we multiply 2 numbers of the form 6k + 1 together, we get another number of the same form (check this!). If 3 there were only finitely many primes of the form 6k + 5, say p0 = 5 < p1 < p2 < ··· < pn: Consider the integer N = 6p1p2 ··· pn + 5: Clearly N > 1 is not divisible by 2 and the note with which we began this solution implies that N has at least one prime divisor p of the form 6k + 5. If p = 5, we get that 5jN − 5 = 6p1p2 ··· pn; while, if p > 5, we have that 5jN − 6p1p2 ··· pn = 5: In either case, we have a contradiction..