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How to show that $16n^2 + 16n+1 \neq m^2$ with $n, m \in \mathbb{N}$ ?

I already know that $m$ needs to be an odd number because:

$16n^2 + 16n + 1 \equiv 1 \mod{4}$

I can complement the square by:

$(4n+1)^2 - 8n \neq m^2$

But from this point on, I dont know how to go on.

thinkingeye
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2 Answers2

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Let $n\in\mathbb{N}^*$, then $$ 16n^2+16n+1=(4n+2)^2-3<(4n+2)^2 $$ and $$ 16n^2+16n+1=(4n+1)^2+8n>(4n+1)^2 $$ Thus $16n^2+16n+1$ can't be a square because if $m\in\mathbb{N}$ is such that $16n^2+16n+1=m^2$, we would have $4n+1<m<4n+2$ which is not. Notice that the case $n=0$ works because $1=1^2$.

Tuvasbien
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$16n^2 + 16n + 1$ reminds me of $4n^2 + 4n + 1 = (2n+1)^2$.

It isn't, but it reminds me. SO if we muck about....

$16n^2 + 16n + 1 = 4(4n^2 + 4n + 1) -3= 2^2(2n+1)^2 - 3= (2(2n+1))^2 -3$.

If we replace $2(2n+1)$ with $K$ and we assume $16n^2 +16n+1 = m^2$ we get:

$16n^2 + 16n + 1 = (2(2n+1))^2 - 3 = K^2 - 3 = m^2$.

$K^2 - 3 = m^2$ and $K^2 - m^2 = 3$ and $(K-m)(K+m) = 3$.

Assuming that $K$ and $m$ are both positive with must have

$K-m = 1$ and $K + m = 3$ which means $m=1$ and $K = 2$ (and $2^2 - 1^2 =3$).

So $2(2n+1) = 2$ or $n =0$

fleablood
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  • Very interesting approach! Maybe this helps me in solving a more general version of this problem. – thinkingeye Apr 18 '20 at 17:56
  • Well if you have $m^2 + c=n$ we will end up with $(m-n)(m+n) = c$. And if we let $n=m+k$ we get $k(m+2k)=c$ which may not have many solutions. ANd if $c$ is prime!..... – fleablood Apr 18 '20 at 18:09