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Assume $n$ is a Natural Number which satisfies the following 2 properties simultaneously:

$01$ . $(3n+1)$=$a$12 for some Natural Number $a$1.
$02$ . $(4n+1)$=$a$22 for some Natural Number $a$2.

Prove that $n$ is divisible by $56$.

For example, if we take $n$=$56$, we have $(3n+1)$=$169$=$13$2 and $(4n+1)$=$225$=$15$2.

I could prove that $n$ is a multiple of $4$ but coud not proceed any further.However my strategy was to prove that $n$ is divisible by 8 & then that $n$ is divisible by 7 because $G.C.D (7,8)=1$ and $L.C.M.(7,8)=56$.

Andrés E. Caicedo
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Adhiraj Mandal
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  • http://math.stackexchange.com/questions/575733/if-4n1-and-3n1-are-both-perfect-sqares-then-56n-how-can-i-prove-this – lab bhattacharjee Jan 11 '14 at 17:00
  • @Ross The (only) answer in the dup you cited has a big error (see the comments). So maybe this should not have been closed. – Bill Dubuque Jan 11 '14 at 17:39
  • I would have thought then that the correct thing is to leave this closed as a duplicate and provide a correct answer to the previous question? – Old John Jan 11 '14 at 18:18
  • @OldJohn Generally I agree. But in this case, the old question now has no answers (2 deleted), so there is no harm in closing it. This, being newer, will get more exposure, so possibly may attract the attention of someone who has the time and knowledge to write a complete, correct answer (the usual approach is to use results about Pell equations, so it will be a little bit of work to write a nice answer). Hopefully someone will find the time since it is a nice problem. – Bill Dubuque Jan 11 '14 at 20:22
  • I wrote an answer to the duplicate question. – Phira Jan 16 '14 at 09:36

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