What went wrong in this fake proof? So, I came up with this quite ridiculous fake proof that no nonnegative integers $(x,y)$ satisfy $x^2+y^2=5$. Clearly, $(x,y)=(1,2)$ satisfies the equation, so the proof is wrong; but somehow, I can't find why!
Let $f(x,y):=x^2+y^2-5$. We will prove by induction on $x$ that $f$ has no roots. Base case: when $x=0$, no $y$ satisfies $f(0,y)=0$ since $y^2=5$, but $\sqrt5$ is irrational (recall that $x,y$ are integers). Now suppose $f(x,y)$ has no roots for some particular $x$. We need to show that $f(x+1,y)$ has no roots too. To do this, we take the contrapositive statement: if $f(x+1,y)$ had a root, say $(a,b)$ then $f(x,y)$ has a root too. This is true, since the solution set $(a+1,b)$ works. Since the contrapositive statement is true, the original statement is true too. This completes induction, since $f(0,y)$ has no roots implies $f(1,y)$ has no roots, etc. such that $f(x,y)$ has no roots for any $x$.
Any help is appreciated!
 A: I think you should really focus on your last sentence, since that shows what goes wrong.  Correctly, there are no integers $y$ with $f(0,y)=5$.  However, as you point out, there is an integer $y$ with $f(1,y)=5$, but your argument says there isn't. 
Your error comes from thinking that you can take a supposed root $y_0$ for $f(1,y)$ and simply move the $x$-coordinate down and preserve the root, thereby showing that $y_0$ is a root for $f(0,y)$.   This isn't so because $f(1,y)=y^2-4$ isn't the same thing as $f(0,y)=y^2-5$.  Or, said differently, the function $f(x+1,y)$ and $f(x,y)$ are completely different.  If you know that $f(x_0,y_0)=0$, why in the world should $f(x_0-1,y_0)$ also equal $0$?  There is no reason.  
What you are likely thinking:  if $(a,b)$ is a root for $f(x,y)$, then $(a+1,b)$ is a root for $f(x-1,y)$.  True, but not what you need here. 
A: You are using induction on the parameter $x$. That's OK for me, but for the fact that I don't see I you could work with it. The problem is, the hypothesis should be "$P(x)$ : $f(x,y)$ has no root (meaning, no $y$ root)".
I think your problem is that you don't define clearly what you understand has a "root" for a multivariate polynomial. Those polynomials may have infinitely many roots... if you call "root" a couple $(x,y)$ such that $f(x,y)=0$.
