$k^{2}+(k+1)^{2}$ being a perfect square for infinitely many $k$ Generally one can see that there are infinite number of solutions for this equation $$a^{2}+b^{2}=c^{2}$$ by taking multiples of the solution $3,4$ and $5$. 
Can i use this as a fact to prove, that $k^{2} + (k+1)^{2}$ is a perfect square for infinitely many $k \in \mathbb{N}$? Any hints, suggestions would be helpful. If not, then how do i prove this fact!
 A: Just to add to Derek's excellent answer,
Pell's equation becomes essential in getting all the solutions to $$k^2 + (k+1)^2 = j^2$$.
Since $gcd(k,k+1) =1 $ we may assume that (by the formula for all primitive pythagorean triples) that
$$m^2 - n^2 = 2mn \pm 1$$
This is a quadratic in $m$ and solving for $m$ gives us
$$ m = n \pm \sqrt{2n^2 \pm 1}$$
Thus any solution of $$k^2 + (k+1)^2 = j^2$$ can be used to get a solution to the equation $$2n^2 \pm 1 = b^2$$ and vice versa.
A: You can find infinitely many by considering solutions of Pell's equation $ m^2 - 2a^2=-1.$
Your (3,4,5) solution comes from$$(7-5\sqrt{2})(7+5\sqrt{2})=-1,$$
where $m=7$ and $a=5$.
For example $$(7-5\sqrt{2})^3$$ yields
$$696^2+697^2 = 985^2.$$
Just to make it a bit more clear note that
$$\left( {m-1 \over 2} \right)^2 + \left( {m+1 \over 2} \right)^2 = a^2.$$
A: This is obvious by ascent in the ternary tree of Pythagorean triples. Explicitly, ascent  yields this formula:  
$\rm (x,x+1,z) \to (X,X+1,Z),\ \ X = 3x+2z+1,\ \ Z = 4x+3z+2\:.\ \ $  For example  
$\rm (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$
This corresponds to always taking the middle branch in the tree, see below
$\qquad\qquad$ 
The reflection that yields the descent in the triples tree is the following
$\quad\quad (x,y,z)\; \mapsto (x,y,z) - 2 \dfrac{(x,y,z)\cdot(1,1,1)}{(1,1,1)\cdot(1,1,1)} (1,1,1)$
$\quad\quad\quad\quad\quad\quad = (x,y,z) - 2 \; (x+y-z) \; (1,1,1)$
$\quad\quad\quad\quad\quad\quad = (-x-2y+2z, \; -2x-y+2z, \; -2x-2y+3z)$
We ascend the tree by inverting this reflection, combined with trivial sign-changing reflections:
$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$
$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$
$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$
Continuing in this way enables one to reflectively generate the entire tree of primitive Pythagorean triples. This has a beautiful geometric interpretation in terms of refelections - see my said MathOverflow post and see here for sums of four squares.
A: It is clear that there is a relationship with solutions of Pell's equation, but it must still be written more clearly and in a more general way.  
Solutions of the equation: $X^2+(X\pm{a})^2=Y^2$   
Defined solutions of Pell's equation: $p^2-2s^2=\pm{a}$   
And the solutions are of the form:   
$X=2s(s+p)$   
$Y=2s(s+p)+p^2$   
These numbers can be different characters.
I almost forgot. If we know what the solution of Pell's equation $p^2-2s^2=a$  
$s_{2}=2p+3s$
$p_{2}=3p+4s$ 
