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Let both $x$ and $y$ be arbitrary naturals and perfect squares. By definition, a perfect square can be written in the form $k^2$, for some natural $k$. So we can write $x = (k+1)^2$ and $y=k^2$, where $k$ is an arbitrary natural. Thus, $x-y=(k+1)^2-k^2=k^2+2k+1-k^2=2k+1.$ By definition, an odd number is an integer that can be written in the form $2j+1$, for some integer $j$. Since, naturals are a subset of integers, then $2k+1$ is odd. Therefore, $x-y$ is odd.

Since this logic works for any integers $x$ and $y$, we have shown that the difference consecutive between perfect squares is odd.

Is this proof correct? If it is correct, is there any way I could improve it?

Robert Shore
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Gabriel Dylan
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    This is absolutely correct. Pretty sure this is how the proof usually goes. – dgeyfman Jul 27 '23 at 23:58
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    Alternatively, you could just note that $x^2$ and $x$ have the same parity, and two consecutive integers have different parity. This has the advantage that it generalizes to other positive integer powers, not just squares. – Robert Israel Jul 28 '23 at 01:13

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