We know that the sum of the odd numbers up to $2n-1$ is $n^2$. The sum of the odd numbers from $2m+1$ through $2n-1$ is then $n^2-m^2=(n+m)(n-m)$ We need this to be $225$, so you are looking for the number of solutions to $225=(n+m)(n-m)$ Each factorization of $225$ gives you one except $225=15\cdot 15$ (Why?) For example $225=45\cdot 5$ We can then write $n+m=45,n-m=5, n=25, m=20$ and find $225=41+43+45+47+49$. Note that there are $5$ terms in the expression and the middle one is $45$
Consider the equation $x^2+y^2=z^2+w^2=N.$
This is equivalent to: $x^2-z^2=w^2-y^2=D.$
and we need $D$ to have at least two different factorizations with the factors having the same parity. One approach is thus to find the smallest values of $D$ which satisfy this property. Another approach is to find parametric solutions.
Let $(x-z)=ab$ and $(x+z)=cd$ with $(w-y)=ac$ and $(w+y)=bd$,
which gives the parametric solution: $\dfrac{1}{2}(ab+cd,bd-ac,cd-ab,ac+bd)$.
We have $N=\dfrac{1}{4}(a^2b^2+c^2d^2+b^2d^2+a^2c^2)$.
Now from the fact that $cd-ab$ and $bd-ac$ are distinct positive integers, we can see that no three of $a,b,c,d$ can be the same.
Thus, the lexicographically smallest value for the tuple $(a,b,c,d)$ is $(1,1,2,3)$, which gives the solution $(7,1,5,5)$ and $N=50$. Once we have this solution, it is easily checked that other tuples of $a,b,c,d$ give larger values of $N$.
Best Answer
If $n^2+(n+1)^2=k^2$, then $2n^2+2n+1=k^2$, so $(2n+1)^2=4n^2+4n+1=2k^2-1$.
So we are looking for solutions to the negative Pell equation $(2n+1)^2-2k^2=-1$,
which are $2n+1=1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319,$
$54608393, ...,$ listed in OEIS.