1
$\begingroup$

I have following expression: $$2n(n+1)$$ Where n is a natural number. We want to know if there exists a subsequence of natural numbers that makes this expression a perfect square. According to existing theorem this expression should be 0 or 1 mod 4. Let's rewrite with $2l$ for $n$. We get: $$8l^2+4l=4(2l^2+l)$$ Which is 0 mod 4. But if we substitute $l=1$ our expression is $12$ which is not a perfect square. Where did I make a mistake?

$\endgroup$
1
  • $\begingroup$ Your mistake is that you are assuming $n$ is even. A fact is that exactly one of the numbers $n$ and $n+1$ is even (so $4\mid n(n+1)$), not necessarily $n.$ $\endgroup$
    – CIJ
    Nov 25, 2016 at 20:59

2 Answers 2

2
$\begingroup$

If $n $ is a perfect square, $n \equiv 0 \mod 4 \vee n \equiv 1 \mod 4$ but $n \equiv 0 \mod 4 \vee n \equiv 1 \mod 4 \not\Rightarrow n $ is a perfect square.

For example $5 \equiv 1 \mod 4$ and 5 is not a perfect square.

$\endgroup$
0
$\begingroup$

It s true that squares are either 0 or 1 mod 4 but the converse is not true as you just showed. Instead, look at primes dividimg 2n(n+1). Notice n and n+1 are always coprime so if an odd prime p divides 2n(n+1), it divides either only n or only n+1. Assuming 2n(n+1) is a square says an even power of p divides either n or n+1. So n and n+1 have to be squares up to factors of 2.

So the cases are $ n=x^2,n+1=2y^2$ or $x=2x^2, n+1=y^2$. This means you are solving $2y^2-x^2=\pm 1$. This does have infinitely many solutions and is a case of Pell's equation.

You can get solutions by looking at powers (1+√2)^k and looking at the coefficients.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .