×

# 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

Sort by:

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 $$5\mid n$$.

- 2 years, 1 month ago

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.

- 2 years, 1 month ago

You could just post it directly to me on Slack.

- 2 years, 1 month ago

Bhai, can you come on Slack ?

- 2 years, 1 month ago

Sure. Give me a few moments.

- 2 years, 1 month ago

Of Course!

- 2 years, 1 month ago