(i) If (x⁴ + x² + 1)² = x⁸+ x⁴ + 1 + 2x⁶ + 2x⁴ + 2x²

then

(x⁴ + x² + 1)² = x⁸ + x⁶ + x⁴ + x²+ 1 + x⁶ + 2x⁴ + x² = x⁸ + x⁶ + x⁴ + x²+ 1 + x²(x² + 1)²

Hence

(x⁴ + x² + 1)² – (x(x² + 1))² = x⁸ + x⁶ + x⁴ + x²+ 1

(ii) If x⁸ + x⁶ + x⁴ + x²+ 1 is prime then its only factors are 1 and the number it represents.

Using the difference of squares factorisation in (i)

x⁸ + x⁶ + x⁴ + x²+ 1 = (x⁴ – x³ + x²- x + 1)( x⁴ + x³ + x²+ x + 1)

So either (a) x⁴ – x³ + x²- x + 1 = 1 or (b) x⁴ + x³ + x²+ x + 1= 1

If (a), then

x⁴ – x³ + x²- x + 1 = 1

x⁴ – x³ + x²- x = 0

x(x³- x² + x – 1) = 0

x(x – 1) (x² + 1) = 0

So x = 0 or x = 1 are the only (real) solutions.

If (b) then

x⁴ + x³ + x²+ x + 1 = 1

x⁴ + x³ + x²+ x = 0

x(x³ + x² + x + 1) = 0

x(x + 1) (x² + 1) = 0

So x = 0 or x = -1 are the only (real) solutions.

If x = 0, then x⁸ + x⁶ + x⁴ + x²+ 1 = 1 which is not prime.

If x = 1, then x⁸ + x⁶ + x⁴ + x²+ 1 = 5 which is prime.

If x = -1, then x⁸ + x⁶ + x⁴ + x²+ 1 = 5 again.

So the only values of x which give the expression a prime value are x = 1 and x = -1, when the expression is equal to 5.

Challenge 276: Make a Prime?