For how many integral values of $R$ is $R^4 - 20R^2+ 4$ a prime number? I tried factorizing but couldn't conclude anything concrete.
Factorizing it, gives $(R^2 - 10)^2 - 96$. What should be my approach now?
HINT $$R^4 - 20R^2 + 4 = (R^2 + 4R - 2)(R^2 - 4R - 2)$$ If this has to be a prime, then atleast one of the factors has to be $\pm 1$. Can you finish it from here?
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Also, what you have written is incorrect. $R^4 - 20R^2 + 4 = (R^2-10)^2 - 96$
$R^4-20R^2+4=R^4-4R^2+4-(4R)^2=(R^2-2)^2-(4R)^2=(R^2-4R-2)(R^2+4R-2)$.Check for the solutions for each factor term=$1$.The integer values of R is the solution set.