When The Sum Of Squares Of Two Consecutive Integers Is Again A Perfect Square? Find all positive integers $n < 200$, such that $n^2 + (n + 1)^2$ is a perfect square.
Well setting this equal to $k^2$ is important. But before that, since all squares $\equiv 0$ or $1$ (mod $3$ $,4)$ so using this we get that one of the two numbers is divisible by $3$ and same for $4$ (one of the two is divisible by $4$). This gives us cases, like if $4,3 |n$ or $4|n$ and $3|n+1$ and so on. however this seems very tedious, and anyway, for example, the first case, I still don't get how I would solve $(12k)^2+(12k+1)^2=m^2$ for $k,m \in \mathbb N$. Help please; I'm stuck.

If we let $m=n+\sqrt{2n^2+(-1)^n}$, starting with $1$, we get a pair of pell numbers that feed directly into Euclid's formula for generating Pythagorean triples which are the ordered pairs (A,B,C) where $A^2+B^2=C^2$.
$$F(m,n):\qquad A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$
Examples:
$$n=1\implies m=1+\sqrt{2+(-1)^1}=1+1=2\quad F(2,1)=(3,4,5)$$
$$n=2\implies m=2+\sqrt{8+(-1)^2}=2+3=5\quad F(5,2)=(21,20,29)$$
$$n=5\implies m=5+\sqrt{50+(-1)^5}=5+7=12\quad F(12,5)=(119,120,169)$$
There an an infinite number of these but if you don't like the alternate $(A<B)$ vs $(A>B)$ you can use this formula, starting with a nonworking $seed$ of $(A_0,B_0,C_0)=(0,0,1)$
$$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2\qquad$$
This produces the triples $\quad(3,4,5)\quad (20,21,29)\quad (119,120,169)\quad ...$
These three are the only triples where $(A,B,C)<200$
