Question

8. (a) Prove that if p and q are prime numbers then p2 + pq is not a perfect square. (b) Prove that, for every integer a and every prime p, if p | a then ged(a,pb) = god(a,b). Is the converse of this statement true? Explain why or why not. (c) Prove that, for every non-zero integer n, the sum of all (positive or negative) divisors of n is equal to zero. 9. Let a and b be integers and let d=gcd(a,b). (a) Use induction to prove that, for every positive integer k, ged(ak, bk) = dk. (b) Use Unique Factorization Theorem or Proposition 6.8.2: GCD From Prime Factorization (GCD PF) to prove that, for every positive in- teger k, gcd(ak, bk) = dk. (c) Explain why, if a and b are coprime, then so are ak and bk for any positive integer k.

Answer 1

p2 + p q = p (p+q) If possible let, b2 + p q = p (p+q) be perfect square. Then, (p+q) must have factor p = p 1 (b+q) - Þ la and since ļ and a q are are primes then þ=q Now, p2 + p q p² + px 2p2, which is never a perfect square. It is a contradiction. Hence, p²+ pq is not a perfect square.

since, pxa, then the common divisor of a gand b and the common divisor of a , bb a, þb remains same Hence, ged (a, a, b) ged (a, pb) ; as ged is the greatest common divisor NO, converse is not troue. Take , then a = b = b ged (h ged(a, bb) ged (+182) = p = ged (PIP) ged (a, b) but pxa is not true (if fact &= a). then then if let nbe non-zero integer if d is a divisor of u (-d) is also dinisor of n. since, u has finitely many divison. then by summing up all divisors as pair (d, nd) the sum of all the divisors of n is equal to Zeno. we get

we know, if du li ki bi PA Bu & where ts b کم min}xniful kn are prime divisors Then ged (a,b) = b min far, El Now, for a positive integer ki , bk) gedlak minh khi, KBY min ka kan, kiny so asko, minh kx, kyl = k = Kemindanny kominfde, BY Pin 5 human capa?) 11 K.minhan, Buy (minka, nes mind, By min fan, Rh P96469,b) y dk ; where daged (a,b).