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 $40$ .

#NumberTheory #Sharky

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 9 months ago

Bhai, can you come on Slack ?

Swapnil Das - 5 years, 9 months ago

Sure. Give me a few moments.

Satyajit Mohanty - 5 years, 9 months ago

@Satyajit Mohanty – Of Course!

Swapnil Das - 5 years, 9 months ago

You could just post it directly to me on Slack.

Sharky Kesa - 5 years, 9 months ago

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 - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$