<s>[INST] $n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$? $n$ is some natural number. Let $x$ be the integer part of $\sqrt n$ and $y$ be the decimal part. If $x^2 - y^2 = 1+4y$ what is $y^x$?
This is some high school problem but I can't solve it. Any help? Thanks.
 [/INST] Since $y$ is the decimal part, we know that $0\leq{y}<1$, then from $x^2=1+4y+y^2$ we get $$1\leq{x^2}<6$$. $x$ is an integer, then $x$ can be $1$ or $2$.
For $x=1$, $y=0$, then $y^x=0$;
for $x=2$, $y=\sqrt{7}-2$, and we have $n=7$ which satisfies the requirement. Then $y^x=11-4\sqrt{7}$.
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