TITLE: Solving Diophantine equation without brute force
QUESTION [1 upvotes]: I am having trouble trying to solve this Diophantine equations:
$$
^3=4^2+4−3
$$
I was wondering if anyone could help me find the integer solutions to these and any advice of techniques to use?
Thank you!

REPLY [0 votes]: Another way:  You can rewrite the equation: 
$$x^3+8=4y^2+4y+5$$
Or $$(x+2)(x^2-2x+4)=4y^2+4y+5$$
Let $(x+2;x^2-2x+4)=d (d\in N^+)$
Note that : $x^2-2x+4=x(x+2)-4(x+2)+12\Rightarrow d\mid 12$
We will prove $4y^2+4y+5\not\mid12$. Assume $4y^2+4y+5\mid12$, then we have $4y^2+4y+5=12k(k\in Z)$
$\Leftrightarrow (2y+1)^2=12k-4=4(3k-1)$
But $(2y+1)^2\equiv0;1\pmod{3}$ and $4(3k-1)\equiv2\pmod{3}$
Can you continue?
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