I have the following equation:
$$ Ω = -2 A^4 - 2 B^4 + 2 C^4 - D^4 + E^4 + F^4 - G^4 - H^4 + I^4 + J^4 - K^4 + 12 A B L P $$
where
- $C = B + A$
- $E = D + A$
- $I = H + G - F$
- $J = H + G - D - A$
- $K = H + G - D$
- $L = H + G - B - A$
- $M = H + G - B$
- $N = H + G - A$
- $P = H + G$
and $A$, $B$, $D$, $F$, $G$, and $H$ are independent variables.
I suspect that the expression can also be represented in the following way:
$$ Ω = a A^4 + b B^4 + c C^4 + d D^4 + e E^4 + f F^4 + g G^4 + h H^4 + i I^4 + j J^4 + k K^4 + l L^4 + m M^4 + n N^4 + p P^4 $$
where $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$, $i$, $j$, $k$, $l$, $m$, $n$, and $p$ are all integer coefficients . However, I don't know how to go about solving a fifteen-variable equation. I suspect the coefficients should all be reasonably small, say within the range $[-6, 6]$, but guess-and-checking even that limited range for that many variables would be wildly impractical to do by hand.
Is there some software or technique that can either find the coefficients or prove that no integer solutions exist?
EDIT
I figured it out for this specific case with the help of some tentative assumptions that luckily turned out to be true and patterns in some alternate equations that turned out to be more related to the one I asked about than I had suspected. All the coefficients are either positive or negative one, giving the following result:
$$ Ω = −A^4 − B^4 + C^4 − D^4 + E^4 + F^4 − G^4 − H^4 + I^4 + J^4 − K^4 + L^4 − M^4 − N^4 + P^4 $$
However, I'd still like to know if there's a good way to approach this problem as presented, without that extra help I had for this specific result, because I have additional similar but distinct equations to work out. For example:
$$ β = \frac{1}{6} (3 A^3 + 2 B^3 - 2 C^3 - D^3 + E^3 + F^3 - G^3 - H^3 + I^3 + J^3 - K^3 + L^3 - M^3 - N^3 + P^3) + A B (H + G - 2 D) - B F^2 + B F G + B F H - B G H $$