Selected solutions to PS 2

Problem 2 Show that any two consecutive perfect squares are relatively prime.

Solution: gcd(n,n+1)=1, so we know that n and n+1 have no common prime factors. The prime factors that occur in n² are the same as those in n, but with twice the multiplicity. Thus n² and (n+1)² have no prime factors in common. Thus gcd(n²,(n+1)²)=1.

Problem 5 Suppose that a and b are relatively prime, and that ab is a perfect square. Show that a and b are perfect squares.

Solution: Because ab is a perfect square, any prime p that occurs in ab must occur with even multiplicity. Now gcd(a,b)=1, so a and b have no common prime factors. Thus every prime that occurs in a must occur with even multiplicity, and similarly for b. Therefore a and b are perfect squares.