A perfect square is an integer that is the square of an integer. So a is a perfe

Question

A perfect square is an integer that is the square of an integer. So a is a perfect square if and only if there exists an integer k so that a = k2.

[2] Prove that for all integers n and m, if n and m are perfect squares then so is m·n.

Suppose n is a perfect square, then there exists an integer k so that n = k^2.

Suppose m is a perfect square, then there exists an integer k so that m = r^2.

nm = (k^2)(r^2) = (kr)^2 where kr is some integer x = (x)^2

Since nm = (kr)^2 where kr is some integer x, by definition of a perfect square n*m is a perfect square. Therefore, if n and m are perfect squares then nm is a perfect square as we

[2] Is the converse true? Give a proof (if it is true) or a counterexample (if it is false)

Not sure how to prove this one. Any help?

Explanation / Answer

Converse - if m*n is perfect square , then m and and n are also perfect squares.

This is Not True.

Let k = m*n = r^2 (let's say k is square of r)

now m and n are factors of k, which may not be necessarily perfect square.

E.g. k = 36, r = 6

m = 12, n = 3 (not perfect squares)

m = 18, n = 2 (not perfect squares)

m = 6, n = 6 (not perfect squares)

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