<s>[INST] Show that the equation $4x=y^2+z^2+1$ has no integer solution 
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
 [/INST] $$4x=y^2+z^2+1$$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 
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