Answer
Claim: Suppose that integers m and n are perfect squares. Then m + n + 2√(mn) is also a perfect square.
Proof: Suppose both m and n are particular but arbitrarily chosen perfect squares.
By definition of perfect square, let m = r² and n = s².
By substitution and algebra, it follows that:
m + n + 2√(mn) = r² + s² + 2√(r² * s²)
= r² + s² + 2 * √r² * √s²
= r² + s² + 2rs
= (r+s)(r+s)
= (r+s)² .
Let t = r+s, then (r+s)² = t².
By definition of a perfect square, it follows that t² = m + n + 2√(mn) is a perfect square.
Work Step by Step
A perfect square is define as y = x². Define m and n as perfect squares. Substitute m and n into the equation m + n + 2√(mn) and use algebra to determine m + n + 2√(mn) = (r+s)². Let t = (r + s), then m + n + 2√(mn) = t². Then conclude that t² is a perfect square.
