Show that the equation $4x=y^2+z^2+1$ has no integer solution

Question

Show that the equation $$4x=y^2+z^2+1$$ has no integer solution.

I divided throughout by $4$ to get $$x=\frac{y^2}{4}+\frac{z^2}{4}+\frac{1}{4}$$ but not sure if that is correct

Answer 1

Since for any $n\in\mathbb{N}$ we have $n^2\pmod{4}\in\{0,1\}$, $Y^2+Z^2+1$ can never be a multiple of four.

Answer 2

$$4x=y^2+z^2+1$$4x is even ,so other side must be even ,so one of y or z is odd and ohter is even
suppose y is odd and z is even $$y=2k+1 \rightarrow y^2=4k^2+4k+1\\$$$$z=2q \rightarrow z^2=4q^2$$put them in furmula $$4x=(4k^2+4k+1)+(4q^2)+1\\4x=4(k^2+k+q^2)+2 $$now divide both side by 2 $$2x=2(k^2+k+q^2)+1$$ and it is impossible ,because left hand side is even ,but right hand side is odd