let x,y be elements of integers. proove that (x+1)y^2 is even if and only if x is odd or y is even.?

Answer 1

mathgirl proved this for x odd y even.

(a) the problem is an "if and only if" proof; and

(b) the condition is an rather than an .

There are three components to this proof: (1) x odd; (2) y even; and (3) the expression is odd (not even) if x is even and y is odd.

Case 1 (x odd): Let x = 2m+1 for some integer m. Then
(x+1)y^2 = (2m+2)y^2 = 2(m+1)y^2 which is even.

Case 2 (y even): Let y = 2m for some integer m. Then (x+1)y^2 = (x+1)(2m)^2 = 4(x+1)y^2 which is even.

Case 3 (x even, y odd): Let x = 2m and y = 2n+1 for integers m and n. Then (x+1)y^2 = (2m+1)(2n+1)^2 = (2m+1)(4n^2 + 4n + 1) = 8mn^2 + 8mn + 2m + 4n^2 + 4n + 1 = 2(4mn^2 + 4mn + m + 2n^2 + 2n) + 1 which is odd (not even).

That completes the proof.

Answer 2

Let x be odd, then for some integer m, x = 2m + 1
Let y be even, then for some integer n, y = 2n
(x+1)y^2 = (2m+1+1)(2n)^2
= (2m+2)(4n^2)
= 8mn^2 + 8n^2
= 2(4mn^2 + 4n^2)
Thus, (x+1)y^2 is even