Find $a,b,c$ if ${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$ If $${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$$
and $$(a+b+c)=21, a\gt6$$
We have to find $a,b,c$
My attempt
I use a little fact that the sum of first $n$ odd numbers is $n^2$
From that i get $$a^2+b^2=c^2$$
Which means that the solutions are pythagorean triples, and found no such soultion.
While if i try to do this by hand, $$a=7,b=5,c=9$$ satisfies the question but not the pythaoarean equation that i set up above
Why is this so?

If $a=2n-1$ so $$1^2+3^2+...+a^2=n^2=\left(\frac{a+1}{2}\right)^2.$$
Id est, $$(a+1)^2+(b+1)^2=(c+1)^2$$ or
$$a^2+b^2+2a+2b+2=(22-a-b)^2$$ or
$$ab-23(a+b)=-241$$ or
$$(a-23)(b-23)=288.$$
Also we have $a\geq7$, $b+c\leq14$, $c>b$ and $c>a$.
Can you end it now?
