TITLE: Prove that there exists rational numbers $x$ such that $\sqrt{x + 2016}$ and $\sqrt{x + 2017}$ are also rational
QUESTION [5 upvotes]: I need to know if my solution is correct. Here is how I did it:
By substituting $x + 2016$ with $t$, our problem reduces to proving that there exists $t \in \mathbb{Q}$ such that $\sqrt t$ and $\sqrt {t + 1}$ are rational numbers.
Let $t = \frac{m^2}{n^2}$, where $m, n \in \mathbb{Z}$. By putting this in $t + 1$, we get $t + 1 = \frac{m^2 + n^2}{n^2}$.
By knowing that $m^2 + n^2 = p^2$, where $m, n, p \in \mathbb{Z}$ has infinite solutions, we are done.

REPLY [3 votes]: Yes, this is correct: $\displaystyle t:=(\frac{m}{n})^2$ => $\displaystyle t+1:=\frac{m^2+n^2}{n^2}$ .
But then set e.g. $m:=a^2-b^2$ and $n:=2ab$ so that you get 
$\displaystyle t=(\frac{a^2-b^2}{2ab})^2$ and $\enspace\displaystyle t+1=(\frac{a^2+b^2}{2ab})^2$ .
An example with positive $x$ :
$a:=90$ , $b:=1\enspace$ => $\enspace\displaystyle  x=\frac{275401}{180^2}$ 
with $\enspace\displaystyle \sqrt{x+2016}=\frac{8099}{180}\enspace$ and $\enspace\displaystyle \sqrt{x+2017}=\frac{8101}{180}$