Find $a,b,c$ if ${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$

Question

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?

Answer 1

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?