I have the following problem to solve: given $P_i$ the set of first 20 primes (that is i=1,2,...20). find m so that $GCD(5+18m,P_i)=1$ ; and $GCD(7+18m,P_i)=1 $ How should I best approach this problem?
2026-04-06 03:36:10.1775446570
Solving set of GCD Equations
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Let $\,P\,$ be the product of all those primes $> 7$ and $\,m = 3P^2$.
If $\,p\neq 5,7\,$ then $\,p\mid 18m,\, p\nmid 5\,$ so $\,p\nmid 5+18m.\,$ Simlarly $\,p\nmid 7+18m.\,$
If $\,p = 5\,$ then $\,5\nmid m\,\Rightarrow\, 5\nmid 5+18m,\,$ and $\,5\mid 7+18m\iff \!\bmod 5\!:\,\ m\equiv -7/18\equiv 3/3\equiv 1$ hence $\, 1\equiv m \equiv 3P^2\Rightarrow\, 3\,$ is a square mod $5$, contradiction. Similarly for $\,p=7.\,$