If $3n + 1$ and $4n + 1$ are perfect squares prove that $n$ is divisible by $56$.
I succeeded in proving that $n$ s divisible by $8$; but i can't prove that it is divisible by $7$.
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3How did you prove divisibility by $8$? – Arthur Sep 29 '17 at 16:26
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$k^2 = 0,1,2 \mod 8$ and $k^2 = 0,1,2,4 \mod 7$ so try some $n$ from $0$ to $55 = 7*8-1$ and see which one gives you a square module with $7$ and $8$ ! – Sep 29 '17 at 16:31
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By congruences you will get it easily also use the fact that the odd squares will be in 4k +1 form – swapnil v Sep 29 '17 at 16:31
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If we are able to disacard the cases 3n+1 congruent 0 mod 7 and 4n+1 congruent 2 mod 7 then we are through i think – swapnil v Sep 29 '17 at 16:34
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https://math.stackexchange.com/questions/575733/if-4n1-and-3n1-are-both-perfect-sqares-then-56n-how-can-i-prove-this – lab bhattacharjee Sep 29 '17 at 16:48