KVPY #15

Let $2a$ , $2b$ and $2c$ be the $n$ th, $p$ th and $q$ th terms of the respective arithmetic progressions .

$\implies \begin{cases} 2a = 4n-2 & \implies n = \dfrac {a+1}2 \\ 2b = 4p-2 & \implies p = \dfrac {b+1}2 \\ 2c = 4q-2 & \implies q = \dfrac {c+1}2 \end{cases}$

Then we have:

$\begin{aligned} (2+4+10+...+2a)+(2+4+10+...+2b) & = (2+4+10+...+2c) \\ (2+4+10+...+(4n-2))+(2+4+10+...+(4p-2)) & = (2+4+10+...+(4q-2)) \\ \frac {n(2+4n-2)}2 + \frac {p(2+4p-2)}2 & = \frac {q(2+4q-2)}2 \\ 2n^2+2p^2 & = 2q^2 \\ n^2+p^2 & = q^2 \\ \left(\frac{a+1}2\right)^2 + \left(\frac{b+1}2\right)^2 & = \left(\frac{c+1}2\right)^2 \end{aligned}$

Assuming $n,p,q$ are pythagorean triples (3,4,5), and since $a>6$ $\implies n = 4$ , $p=3$ and $q=5$ and then:

$\begin{cases} a = 2n - 1 = 2(4) -1= 7 \\ b = 2p-1 = 2(3) - 1 = 5 \\ c = 2q-1 = 2(10)-1 = 9 \end{cases}$

Note that $a+b+c = 7+5+9 = 21$ as given. Therefore, $a-b = 7-5 = \boxed{2}$ .