Using remainder theorem in Pythagoras theorem makes absurd results! At first, I apologize for the title. I really couldn't find anything better than this.
Now,we know, some integers $a,b,c$ (none of them are $0$) can be found so that $$a^2 = b^2 + c^2$$ 
Now,here,of course $a>b$. 
So, let, $$a = b + k $$where $k$ is an integer.
So, $$(b+k)^2=b^2+c^2\\\Rightarrow b^2 + 2bk +k^2 = b^2+c^2\\\Rightarrow2b=\frac{c^2-k^2}{k}$$
Now, let, $$c^2-k^2=f(k)$$ 
So, as $b$ is an integer, $f(k)$ should be divisible by $$k=k-0$$ 
So, according to remainder theorem, $f(0)$ should be $0$. 
BUT, $$f(0)=c^2-0^2=c^2\neq 0 (!!!)$$ 
which also concludes that there can beno $a,b,c$ for which $a^2=b^2+c^2 $  (!!!)
I really can't understand what I have done wrong here?? Some help would be highly appreciated.
 A: Your mistake is in saying that $f(k)$ is defined for every $k$. You cannot say that.
If you define $f$ as $f(k)=c^2 - k^2$, you can only do that if you already fixed the value of $c$, and once you fix $c$, the function $f(k)$ is determined for each value of $k$. In fact, then, $f(0)$ is equal to $c^2$ by definition.
On the other hand, if you start from the pythagorean triples, you can also say that $f(k)$ is defined as $c^2 - k^2$ where you pick $c$ in such a way that there exists such an integer $b$ such that $(b+k)^2 + b^2=c^2$. But in this case, 


*

*you have no basis on which to claim that the function $f$ would be equal to $c^2 - k^2$ for some fixed $c$ (in fact, you have proven that this is not the case)

*you don't even know if $f$ is well defined for all values of $k$, and it is in fact not defined for certain values. It most certainly is not defined for $k=0$, for example.

A: You should have determined the domain of the function prior to testing its value for any specific argument!
$f(0)$ is simply undefined – it does not exist, as $k=0$ means $a=b$, which implies $c=0$ and that contradicts your initial assumptions.
