Out of Nowhere?

Gelfond allegedly established the following identity for any positive real number $a$ , $b$ , and $c$ , and $n= 0, 1, 2, 3, 4, 5$ .

$a^n + (a+b+4c)^n + (a+2b+c)^n + (a+4b+9c)^n + (a+5b+6c)^n + (a+6b+10c)^n = (a+b)^n + (a+c)^n + (a+2b + 6c)^n + (a+4b+4c)^n + (a+5b + 10c)^n + (a+6b+9c)^n$

Note that all of the terms in the equation are given as integers. However, it could be that some of $a$ , $b$ , and $c$ may be non integers. Nonetheless, they must still produce integers at all the terms of the expression.

That should be the case if the denominators would be a factor of $180$ ( $\text{lcm}(4,5,9)$ ), with numerators greater than it, since this will yield integers in all the terms of the equation. However, this, too, is only plausible when none of the terms involve a single $a$ , or $b$ , or $c$ . Since none of these conditions can be applied to the equation, we resort to the condition that all of $a$ , $b$ , and $c$ are integers.

So, we choose any three terms on the LHS of the general equation, and equate them to any of the three terms on the LHS of the given equation, to give us a system of equations in $a$ , $b$ , and $c$ . If any among $a$ , $b$ and $c$ become non integers, then we may have mismatched the expressions on the general equation to the given.

For instance, if we let $a=9$ , and we assign $a+b+4c = 28$ , and $a+2b+c = 33$ , we will get $b=11$ and $c=2$ .

Now, we can substitute these for all the terms at the right hand side, giving us

$A = 20, B= 11, C= 43, D=61, E=84, F=93$

That gives us $20\times 11 \times 43 \times 61 \times 84 \times 93 = \boxed{4507992720}$ .

This is now the identity we have come up with, for $n=0,1,2,3,4,5$ .

$9^n + 28^n + 33^n + 71^n + 76^n + 95^n = 20^n + 11^n + 43^n + 61^n + 84^n + 93^n$