Prove that there are at least $2005$ pairs of $(x, y)$ of non negative integers x and y that satisfy $x^2+y^2 = N,$ for some positive integer $N$. Prove that there exists a positive integer $N$ such that there are at least $2005$ ordered pairs $(x,y),$ of non-negative integers $x$ and $y,$ satisfying $x^2 + y^2 = N.$
I have that the $sqrt(N)$ would be the lcm of the hypotenuses of the first $2005$ Pythagorean triples. Since there are infinite amounts Pythagorean triples, there would be $2005$ pairs of $(x, y).$ But, I don't know if this proves what the question asks for completely.
 A: In this paper you can see that the number $r_2(n)$ of integral solutions to $x^2+y^2=n$ is exactly $4(d_1(n)-d_3(n))$ (provided that $n$ can be written as a sum of two squares) where  $$d_i(n)=\sum\limits_{\substack{d\mid n\\d\equiv i\pmod{4}}}1$$ Now if we take $n$ to be a product of primes of the form $4m+1$, then $d_3(n)=0$ and $r_2(n)=4d_1(n)$. Hence $r_2(N)\geq2005$ for sufficiently large $N$.
A: 
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica code:
In[1]:=ParallelTable[
 If[TrueQ[Length[Solve[{x^2 + y^2 == n}, {x, y}, PositiveIntegers]] >=
     2005], n, Nothing], {n, 1, 10^7}]

In the code, I am looking for the number of solutions that the command Solve[] finds, by using the command Length[]. Than I only return a solution for $\text{n}$ that satisfy the condition that the number of solutions is bigger or equal to $2005$, where I solved the equation $x^2+\text{y}^2=\text{n}$ for positive integers $x$ and $\text{y}$.
Running the code gives:
Out[1]={}

This shows that there are no solutions, to your problem, where $1\le\text{n}\le10^7$. 

Note: this does not imply that there is no solution but only that there is no relatively 'small' solution.

A: Consider the line $\ell$ through $(0,1)$ and $(p/q,0)$, where $p/q\in\mathbb{Q}$ is a rational number greater than $1$ written in lowest terms. $\ell$ intersects the unit circle at ${\left(\frac{2pq}{p^2+q^2},\frac{p^2-q^2}{p^2+q^2}\right)}$, which corresponds to the Pythagorean triple $(2pq,p^2-q^2,p^2+q^2)$. In particular, if $q=1$ and $p$ is even, these form the points ${\left(\frac{2p}{p^2+1},\frac{p^2-1}{p^2+1}\right)}$.
So, you can create as many rational points on the unit circle as you want, say $2005$ of them. Multiply each solution by the common denominator of each point and you're done.
