Prove that for all $n\in\mathbb{N}$ the number $\sqrt{n(n+1)}$ is irrational.
My first move would be:
Let's assume that it's not, that it $\sqrt{n(n+1)} = \frac{a}{b}$, where $a,b\in\mathbb{N}$ and $a,b$ are coprime. Then: $$ n(n+1)=\frac{a^2}{b^2} \iff a^{2}=n(n+1)b^2 $$ I do not know how to continue this proof (I suppose that it is worth to notice that $n(n+1)$ is the divisor of $a^2$). Thanks in advance for any help!
On
Hint: $n(n+1)$ is even, so the right-hand side is even. Thus $a$ must be even, so write it as $a = 2 a'$ ($a'$ may or may not be even, but is an integer), giving $4 a'^2 = n(n+1) b^2$. What can you now say about $b$ or $n(n+1)$? Maybe write $N = n(n+1)$, and then write $N = 2^k l$ for integers $k$ and $l$.
Note that $n(n+1)$ is a perfect square if $n=0$. So we assume $n\gt 0$.
Suppose to the contrary that $\sqrt{n(n+1)}$ is rational. Then as in the OP there exist relatively prime integers $a$ and $b$, with $b$ positive, such that $a^2=n(n+1)b^2$.
If $b\gt 1$, then there is a prime $p$ that divides $b$. Then $p$ divides $a^2$, so $p$ divides $a$, contradicting the fact $a$ and $b$ are relatively prime.
Thus we must have $b=1$. Now we show that $n(n+1)$ cannot be a perfect square if $n\gt 0$. If $n^2+n$ is a perfect square, so is $4n^2+4n$. But $4n^2+4n=(2n+1)^2-1$, and the only two consecutive perfect squares are $0$ and $1$.