Using the conclusion of the proof that if $a\in\mathbb{Z}$ then a is even if and only if $a^2$ is even which is:

$$a^2 = (2j-1)^2 = 4j^2-4j + 1 =2(2j^2-2j) + 1 = 2K+ 1;K\in\mathbb{Z}$$

How do I Prove that an even perfect square cannot have the form $4k+ 2$ for $k \in\mathbb{Z}$?

Any help to point me in the right direction would be greatly appreciated!

  • 1
    $\begingroup$ Hint: factor out a "2" from the expression. The other number is even or odd? $\endgroup$ – Phicar Oct 27 '16 at 3:10
  • $\begingroup$ What you have written in your post is a sketch of the proof that the square of an odd number is odd. But what will be more helpful is to assume that $a$ is even, hence write it in a certain way, and then square that expression, and see what you get. $\endgroup$ – tracing Oct 27 '16 at 4:58
  • $\begingroup$ Hint $\ x^2$ even $\,\Rightarrow\, x\,$ even $\,\Rightarrow\,4\mid x^2,\,$ but $\,4\nmid 4k+2\ \ $ $\endgroup$ – Bill Dubuque Oct 27 '16 at 18:09

$4k+2 = 2 (2k+1)$. Since $2k+1$ is odd it doesn't have a factor of 2, so you cannot even write $4k+2$ as a product of two even numbers (so it can't possibly be a square of 2 even numbers).


Let the square root be $x$, we know that it is even.

If $x \equiv 0 \mod 4$, then $x^2 \equiv 0 \mod 4$.

If $x \equiv 2 \mod 4$, then $x^2 \equiv 4 \equiv 0 \mod 4$.


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