Find all possible integers $(a,b,c,d)$ such that $$\sum_{\sigma}\frac{1}{a}=\frac{2018}{abcd}$$ $$abcd+\sum_{\sigma} ab -\sum_{\sigma} a +3=0$$.
It factorises into $\prod_{a,b,c,d} (1-a)=-2018$ and then there are a lot of cases which I can't handle.
2026-03-31 03:24:50.1774927490
Solution to equations
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The first condition gives $$\sum_{cyc}abc=2018.$$
Thus, $$1-\sum_{cyc}a+\frac{1}{4}\sum_{sym}ab-\sum_{cyc}abc+abcd=1+2015-2018$$ or $$(1-a)(1-b)(1-c)(1-d)=-2$$ and we got not so many very strange cases.
How we can get $\sum\limits_{cyc}abc=2018?$
For your new problem use that $1009$ is a prime number.