Prove that if $x$ is odd then $x^2 + (x + 2)^2$ is divisible by $2$ but not for $4$

Question

This is what I tried:

$$x^2 + x^2 + 4x + 4=2x^2 + 4(x + 1),$$

so it's divisible by $2$, since this expression is a sum of a multiple of $2$ and a multiple of $4$. Therefore, for the expression not to be a multiple of $4$, $2x^2$ can't be multiple of $4$, that's what I can't prove. Brazilian student, sorry for my English.

It's the sum of an odd multiple of $2$ and a multiple of $4$. — Chappers, Apr 22 '17 at 01:05
Yep, now I noticed. x^2 should be divisible by 2, so that, when multiplied by 2, would be a multiple of 4. Thanks bro! That's what I didn't think — Vinicius Guimarães, Apr 22 '17 at 01:09
if $x$ is odd, then $x^2$ is odd. Alternatively, replace $x$ with $2n+1$ — Doug M, Apr 22 '17 at 02:10

score 3 · Answer 1 · answered Apr 22 '17 at 01:07

Perhaps the easiest way is to put $x=2k-1$, $k$ an integer, since $x$ is odd. Then $$ x^2+(x+2)^2 = (2k-1)^2+(2k+1)^2 = 8k^2+2, $$ which is clearly $2$ more than a multiple of $4$, and hence not a multiple of $4$ (but is $2(4k^2+1)$, so obviously a multiple of $2$).

score 0 · Answer 2 · answered Apr 22 '17 at 09:34

0

HINT:

$$x^2+(x+2)^2=\{(x+2)-x\}^2+2x(x+2)$$

As $x$ is odd $\iff x+2$ is odd, so will be $x(x+2)$

answered Apr 22 '17 at 09:34

lab bhattacharjee

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Prove that if $x$ is odd then $x^2 + (x + 2)^2$ is divisible by $2$ but not for $4$

2 Answers2