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It is given: $3m+1=$perfect square. Express $m+1$, as the sum of $3$ perfect squares.

I tried to solve the problem by checking for odd and even values perfect squares $4k^2,{(2k+1)}^2$. I got somewhat convincing form of result from $3m+1={(2k+1)}^2$:

$m+1=\dfrac43(k^2)+\dfrac43(k)+1,$ so that $k$ must be a multiple of $3$, but I could not find the exact value of $k$, so that RHS is a perfect square.

as for the part $3m+1=4k^2$, I am hopeless. Help please.

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  • $\begingroup$ Duplicate of this $\endgroup$ – Calvin Lin May 14 '14 at 6:53
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$3m + 1 = n^2$. This means: $n = 3k - 1$ or $n = 3k + 1$. If $n = 3k + 1$ then:

$m = \dfrac{n^2 - 1}{3}$, so $m + 1 = \dfrac{n^2 + 2}{3} = \dfrac{(3k + 1)^2 + 2}{3} = \dfrac{9k^2 + 6k + 3}{3} = 3k^2 + 2k + 1 = k^2 + k^2 + (k+1)^2$.

The other case is similar.

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Hint

$$ 3m + 1 = (3k \pm 1)^2 = 9k^2 \pm 6k + 1 $$ implies $$ m + 1 = \quad ? $$ Now consider $$ k^2 + k^2 + (k \pm 1)^2 $$

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