Is there a non-numerical way to find the positive integer solutions of this equation: $(591-b-c-d)^2+(591-b-c-d)+41 = bcd$ ?
Numerically I know the answer is $b=47,c=53,d=71$ or a permutation of these values.
Any help would be appreciated.
2026-03-25 07:43:29.1774424609
Positive integer solutions of $(591-b-c-d)^2+(591-b-c-d)+41 = bcd$
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The hint.
Show that $a^2+a+41$ is a prime number for all $2\leq a\leq39.$
Also, $a=40$ and $a=41$ they are not valid and for some numbers $a>41$ we get that $a^2+a+41$ is also prime.
Also, by AM-GM $$a^2+a+41\leq\left(\frac{591-a}{3}\right)^3,$$ which gives $a\leq422.$