<s>[INST] Prove that $x^2-5y=3$ has no integer solutions. Here is my attempt:
$x^2-5y=3$ is the same as $2x^2-10y=6$, which is equivalent to $2x^2=10y+6$. 
This means that 2 divides $x^2$. Since 2 is prime, then 2 divides x, or $x=2k$ for some integer $k$.
Then we have that $(2k)^2-5y=3$, or $4k^2=5y+3$. This implies that 2 divides $5y+3$, but this is a contradiction. 
This looks valid to me but any feedback is greatly appreciated.
 [/INST] Hint: $x^2 - 5y = 3 \iff x^2 = 5y + 3$.  Now observe that the units digit of a perfect square must be one of the following digits:  $0, 1, 4, 5, 6, 9$.  
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