# Prove that n(n+1) is never a square.

1. Jun 29, 2014

### cragar

1. The problem statement, all variables and given/known data
Prove that n(n+1) is never a square for n>0
3. The attempt at a solution
n and n+1 are relatively prime because if they shared common factors then it should divide their difference but (n+1)-n=1 so 1 is their only common factor.
So the only possible way for n(n+1) to be a square is if n and n+1 are squares.
I will show that it is impossible to have perfect squares that are 1 apart.
Let x^2 and y^2 be squares that are 1 apart
so we have $x^2-y^2=1=(x+y)(x-y)$
if x>y then x-y is at least 1 and x+y is bigger than 1 so this cant happen.

2. Jun 29, 2014

### Zondrina

You wish to show:

$n^2 + n ≠ a^2, \quad a \in \mathbb{N}, a > 0$

Suppose by contradiction that $n(n+1)$ is a square. Then:

$n^2 + n = a^2$
$0 = a^2 - n^2 - n$

It must be the case that this equation is reliant on the relationship between $a$ and $n$. That is, if $a = n$, then the equation is zero only if $n = 0$, which is a contradiction. What if $n ≠ 0$? Look for two more contradictions.

3. Jun 29, 2014

### AlephZero

Your proof is OK, but it might be easier to use the fact that $n^2 < n(n+1) < (n+1)^2$.