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Let $m$ and $r$ be an even, and $n$ be odd

Let the following be a square defined as the sum of two even numbers

$$m^2 = 10 n + r $$

Examples that would satisfy this equation:

$$ \begin{array} {c|rrrr} \\ \hline m & 4 & 6 & 14 & 16 & 24 & 26 & 34 & 36\\ n & 1 & 3 & 19 & 25 & 57 & 67 & 115 & 129 & \\ \hline m^2 & 16& 36& 196& 256& 576& 676& 1156& 1296 \\ &10(1)+6& 10(3)+6& 10(19)+6&10(25)+6&10(57)+6&... \end{array} $$

$r$ appears to always be 6. I would like to prove that in this pattern, r must be 6 in order to satisfy $m^2 = 10n + r$, but I cannot find a way. It is likely that there is a relationship between $m$ and $n$ but I have not been able to derive that. I have noticed that $ \lfloor \frac {m^2}{10} \rfloor = n$

How could I go about proving that $r$ must be 6?

lambda
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  • Well an even square is divisible by 4 but 10odd is not so k is not divisible by 4. 2 is never the last digit of a square. 6 is the only possible option. But if 10 doesn't require an odd coefficients this wouldn't be the case. – fleablood Oct 12 '16 at 06:47

2 Answers2

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You are restricting that the last digit of $m$ must be either $4$ or $6$, apart from the fact that $m$ is even. Note that for $m=8$, the above is not true.

Let $m$ be any even number. Then, we have $m=2k$, where $k\in\mathbb Z$. Thus, $m^2=4k^2$. As a result, $$m\equiv0\pmod 4$$

If the last digit is $4$ or $6$, then the last digit of $m^2$ must also be $6$. Thus, $$m^2-6\equiv0\pmod{10}$$

Moreover, as $m\equiv0\pmod4$, the multiple of $10$, $n$, must be odd. Suppose this was not true. Then, let $n=2r$, where $r\in\mathbb Z$. It follows that $20r\equiv0\pmod4$. But $m^2=20r+6\equiv0\pmod4$ and $6$ is not divisible by $4$, which is a contradiction.

Thus, $r=6$.

GoodDeeds
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$4|m^2=10n+r $. $4\not \mid 10n $ so $4\not\mid r $. So $r=2,6$. But $r $ is last digit of a square and only $0,4,6$ can last digit of squares.

So $r=6$.

fleablood
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