4
$\begingroup$

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?

  • 2
    You didnt factor it, to factor it, find its zeros.2012-06-29

2 Answers 2

7

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?

Move your mouse over the gray area for a complete answer.

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

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

  • 2
    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.2012-06-29
2

$R^4-20R^2+4=R^4-4R^2+4-(4R)^2=(R^2-2)^2-(4R)^2=(R^2-4R-2)(R^2+4R-2)$.Check for the solutions for each factor term=$1$.The integer values of R is the solution set.