Every odd integer is of the form 2n+1 where n is an integer.
Furthermore if n is any integer, 2n+1 is an odd integer.

Thus the square of any integer is 

(2n+1)² = (2n+1)(2n+1) = 4n^2+2n+2n+1= 4n^2+4n+1 = 2(2n+2)+1.

And since 2n+2 is an integer, say p, 2(2n+2)+1 = 2p+1, which
is an odd integer.

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To use that to prove the proposition:

"if p² is even, then p is also even",

we use what is called "reductio ad absurdum", or indirect proof:

We assume, for contradiction, that the proposition is false.
Then we are assuming that p² is even but p is odd.
But if p is odd then, by the above theorem, p² is odd.  
But p² is even, so we have reached a contradiction.

Therefore the assumption that p is odd is false, and therefore p is even.

Edwin