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Homethe difference of two odd squares is always a multiple of 8

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# the difference of two odd squares is always a multiple of 8

Theorem. Given any odd real integers $m$ and $n$, the difference of their squares is always a multiple of 8. That is, $8|(m^{2}-n^{2})$.

For example, given $m=9$ and $n$ varied from $-1$ to 13 in steps of 2, calculating $m^{2}-n^{2}$, we get the sequence 80, 80, 72, 56, 32, 0, $-40$, $-88$.

Proving this theorem was one of the problems in the 1851 math exam Lewis Carroll took. The difficulty of this problem is one of perception: so often a simple-sounding theorem in number theory turns out to have a rather complicated proof. But for this simple theorem, the proof turns out to be quite simple, almost bordering on triviality.

###### Proof.

Since it was stipulated that $m$ and $n$ are both odd, we can rewrite them as $m=2a+1$ and $n=2b+1$. The square of $m$ is then $m^{2}=(2a+1)(2a+1)=4a^{2}+2a+2a+1$, which simplifies to $4a^{2}+4a+1$. Likewise, $n^{2}=4b^{2}+4b+1$. Their difference is then $(4a^{2}+4a+1)-(4b^{2}+4b+1)=(4a^{2}+4a)-(4b^{2}+4b)$. This can’t be reduced to fewer terms. However, either side can be rewritten so as to not explicitly use squaring: $4a^{2}+4a=4a(a+1)$. This reveals that either the left-hand or right-hand side of our subtraction is a number that is four times a pronic number, that is, a number of the form $a(a+1)$. All pronic numbers are even, but at this point we can’t distinguish between singly even numbers and doubly even numbers. But, as it turns out, each pronic number is twice a triangular number, which is an integer. So $4a(a+1)$ is eight times some triangular number $T_{c}$, and $4b^{2}+4b=8T_{d}$. We rewrite our subtraction yet again: $m^{2}-n^{2}=8T_{c}-8T_{d}$. By redistributing, we get $m^{2}-n^{2}=8(T_{c}-T_{d})$, proving the theorem. ∎

## Mathematics Subject Classification

11-00*no label found*30-00

*no label found*26-00

*no label found*

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