Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.

Question

Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.

First, I took $$2n+1 \equiv x^2 \equiv 0, 1 \pmod 4$$ which showed that $n$ was even.

Now, $$3n + 1 \equiv y^2 \equiv 0, 1, 4 \pmod 8$$ But since $n$ is even, we get that $8|n$.

So, now any square $\equiv 0, 1, 4, 5, 6, 9 \pmod{10}$. So, I tested $2n+1$ and $3n+1$ for all numbers from 0 to 9. For only two, 0 and 5, were both of them ending with the residues above mentioned. So, I finally proved that $5|n$.

So, $40|n$.

Is my proof correct?

Also, my proof is too roundabout and lengthy. I had to write many programs to take different modulos. Can anyone suggest a more elegant proof especially for the second part when I have to show that $5|n$?

Thanks.

This is a Putnam question, which has the solution "Think modulo $5$ and modulo $8$". — Dietrich Burde, May 07 '16 at 15:55
The proof looks good to me. One thing to make it a little easier is to note that once you've shown that $n$ is even, showing $5|n$ is sufficient. The only squares mod $5$ are $0,1,$ and $-1$. — Kevin Long, May 07 '16 at 15:55
If $8|n$ and $5|n$, then $40|n$. There's no reason to show $10|n$ if you already have that $n$ is even. — Kevin Long, May 07 '16 at 16:05
Possible duplicate of Positive integer $n$ such that $2n+1$ , $3n+1$ are both perfect squares. — Dietrich Burde, May 07 '16 at 16:17

score 6 · Answer 1 · answered May 07 '16 at 16:00

You basically have the idea - note that the quadratic residues $\mod{5},\mod{8}$ are both $\{0,1,4\}$, and you need to find $n$ such that $(2n+1),(3n+1)$ are both amongst these for each modulus.

Checking cases gives that $(2n+1)\mod{5}$ and $(3n+1)\mod{5}$ are both in $\{0,1,4\}$ only when $n=0\mod{5}$, and the same calculations $\mod{8}$ tell you that $n$ must be $0\mod{8}$.

Thus, $n$ is a multiple of $5$ and $8$, which are coprime, hence $n$ is a multiple of $5\times8=40$.

score 3 · Accepted Answer · answered May 07 '16 at 16:47

Your proof is good.

I suppose you could be more direct by pointing out:

For any natural $m$, $m^2 \equiv 0,1,4 \mod 8$ and $m^2 \equiv 0,1,-1 \mod 5$.

So $2n+1$ and $3n+1$ being perfect squares implies:

a) $2n \equiv -1, 0 ,3 \mod 8$.

As $2n$ is even $2n \equiv -1, 3 \mod 8$ are impossible so $2n \equiv 0 \mod 8$ so $n \equiv 0, 4 \mod 8$.

b) $3n \equiv -1, 0, 3 \mod 8$.

As $n \equiv 0, 4 \mod 8$, $3n \equiv 0, 4 \mod 8$ (respectively). As $3n \equiv 4 \mod 8$ isn't possible $n \equiv 4 \mod 8$ isn't possible so $n \equiv 0 \mod 8.$

c) $2n \equiv -1, 0, -2 \mod 5 \equiv 4, 0, -1 \mod 5$.

So $n \equiv 2, 0, -1 \mod 5$

d) $3n \equiv -1, 0, -2 \mod 5 \equiv -6, 0, 3 \mod 5$

So $n \equiv -2, 0, 1 \mod 5$

The only compatibility between c) and d) is $n \equiv 0 \mod 5$

So we have $8|n$ and $5|n$ so $40|n$.

Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.

2 Answers2