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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.

George R.
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    Although correct, I think it'd be nice to give an explicit example of such $t$. – Wojowu Nov 25 '16 at 15:09
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    How about $x=-2016$? – barak manos Nov 25 '16 at 15:13
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    The question gets a little more interesting by requiring $x\ge 0$. That avoids the trivial solution, but all one needs is a Pythagorean triple with $m\gg n$. Such can be obtained from $(\frac 35+i\frac45)^N$ because these are dense in $S^1$. – Hagen von Eitzen Nov 25 '16 at 15:15

2 Answers2

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It's the correct proof. But minimal correct proof is (x = -2016)

kotomord
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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 $\enspace\displaystyle \sqrt{x+2017}=\frac{8101}{180}$

user90369
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