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A Strange Number Theory proof?

There exists positive integers \(n\) such that \(2n + 1\) and \(3n + 1\) are perfect squares. Prove that \(n\) is divisible by \(40\).

Note by Sharky Kesa
2 years, 1 month ago

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If \(n\) is odd, then \(2n+1=2(2k+1)+1=4k+3\). But \(3\) is not a quadratic residue mod \(4\).

If \(n\equiv \{2,4,6\}\pmod{8}\), then \(3n+1\equiv\{7,5,3\}\not\equiv x^2\pmod{8}\). Therefore \(8\mid n\).

If \(n\equiv\{1,3\}\pmod{5}\), then \(2n+1\equiv\{3,2\}\not\equiv x^2\pmod{5}\).

If \(n\equiv\{2,4\}\pmod{5}\), then \(3n+1\equiv\{2,3\}\not\equiv y^2\pmod{5}\). So \(5\mid n\).

Mathh Mathh - 2 years, 1 month ago

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Actually \(n\) is divisible by 40 as well. Try it. I've a solution, but I'm awaiting others to post theirs. Meanwhile, I'd say that the problem would have been much accurate had you asked us to prove that \(n\) is divisible by 40 instead of 8.

Satyajit Mohanty - 2 years, 1 month ago

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You could just post it directly to me on Slack.

Sharky Kesa - 2 years, 1 month ago

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Bhai, can you come on Slack ?

Swapnil Das - 2 years, 1 month ago

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Sure. Give me a few moments.

Satyajit Mohanty - 2 years, 1 month ago

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@Satyajit Mohanty Of Course!

Swapnil Das - 2 years, 1 month ago

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