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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?

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You didnt factor it, to factor it, find its zeros. –  user1708 Jun 29 '12 at 8:03
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2 Answers

up vote 7 down vote accepted

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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$$(R^2 + 4R - 2) = 1 \implies R^2 + 4R - 3 = 0 \implies R \notin \mathbb{Z}$$ $$(R^2 + 4R - 2) = -1 \implies R^2 + 4R - 1 = 0 \implies R \notin \mathbb{Z}$$ $$(R^2 - 4R - 2) = 1 \implies R^2 - 4R - 3 = 0 \implies R \notin \mathbb{Z}$$ $$(R^2 + 4R - 2) = -1 \implies R^2 + 4R - 1 = 0 \implies R \notin \mathbb{Z}$$ Hence, $R^4 - 20R^2 + 4$ is not a prime for all $R \in \mathbb{Z}$.

Also, what you have written is incorrect. $R^4 - 20R^2 + 4 = (R^2-10)^2 - 96$

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Sir, it gives R² - 4R - 3 = 0 and R² + 4R - 3 = 0 But both of them have irrational roots. So the answer will be no solution. –  Bazinga Jun 29 '12 at 7:59
    
Thank you sir, got my mistake. –  Bazinga Jun 29 '12 at 8:02
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Yes. You also need to check one of the factors being $-1$, since both the factors can be negative which will result in a positive number. –  user17762 Jun 29 '12 at 8:03
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$R^4-20R^2+4=R^4-4R^2+4-(4R)^2=(R^2-2)^2-(4R)^2=(R^2-4R-2)(R^2+4R-2)$.Check for the solutions for each factor term=$1$.The integer values of R is the solution set.

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