<s>[INST] How do i find all integers $y$ such that $y^3 = 3x^2+3x+7$, where $x$ is also an integer? I have tried to find all integers $y$ such that $$y^3 = 3x^2+3x+7$$, where 
$x$ is also an integer but i didn't succed only i guess that no integer $y$ 
$x$satisfied that equation so i would like to solve to find all integers 
satisfiing it ?
Thank you for any help .
 [/INST] Try to take mod $3 \rightarrow y^3 \equiv 1 \pmod {3} \rightarrow y \equiv 1 \pmod {3} \implies y^3 \equiv 1 \pmod {9}$ 
Therefore $3x^2 + 3x + 7 \equiv 1 \pmod {9} \rightarrow 3x^2 + 3x \equiv 3 \pmod {9} \rightarrow x^2 + x \equiv 1 \pmod {3}$
This is impossible. If $x \equiv 1 \pmod {3}, x^2 + x \equiv 2 \pmod {3}$ If $x \equiv 2 \pmod{3}, x^2 + x \equiv 0 \pmod{3}$ and obviously if $x \equiv 0 \pmod {3}, x^2 + x \equiv 0 \pmod{3}$
Therefore no such x and y exist where $x$ and y are both integers
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