Calculate 'interference' of number patterns I have 2 numerical series like this:
$$
144 + 25 + 27 + 29 + 31 + \cdots
$$
$$
133 + 3 + 5 + 7 +9 +11+13+\cdots
$$
Is there a efficient way to find the common sum of these patterns?
solution for this case:
$$
144 + 25 + 27 = 133 +3+5+\cdots+15$$
 A: You can see
$$144+\sum_{k=1}^{m}\{25+2(k-1)\}=133+\sum_{k=1}^{n}\{3+2(k-1)\}$$
$$\iff 144+23m+m(m+1)=133+n+n(n+1)$$
$$\iff m^2+24m+144=n^2+2n+133$$
$$\iff n^2+2n-m^2-24m-11=0$$
$$\Rightarrow n=-1+\sqrt{m^2+24m+12}$$
Now we need to know when $m^2+24m+12$ is a square number, so let it be $k^2$.
$m^2+24m+12=k^2\Rightarrow m=-12+\sqrt{132+k^2}.$ Hence, this gives us 
$$132+k^2=l^2\iff (l-k)(l+k)=132$$$$\Rightarrow (l-k,l+k)=(1,132),(2,66),(3,44),(4,33),(6,22),(11,12)$$
Here, since $k,l\in\mathbb N$, $l-k$ and $l+k$ have to have the same parity. (Otherwise, $l,k$ aren't natural numbers. For example, you'll get $(l,k)=(133/2,131/2)$ from $(l-k,l+k)=(1,132)$. This has to be eliminated.)
So, only two pairs $(l-k,l+k)=(2,66),(6,22)$ are left. Then, $(l,k)=(34,32),(14,8).$
Hence, noting that $m=-12+\sqrt{132+k^2}=-12+l$, we get $m=2,22$. So, $(m,n)=(2,7),(22,31).$
Therefore, we now know that there are only two examples as the followings :
$$144+25+27=133+3+5+\cdots+15$$
$$144+25+27+\cdots+67=133+3+5+\cdots+63.$$
