Fundamental of Mathematics 2

n⁴+4 ( add & subtract 4n² )

n⁴+4+4n²-4n²

(n²+2)²-(2n)²

(n²+2+2n)(n²+2-2n)

This is prime so

(i) n²+2n+2=1

From here n=-1 & n⁴+4=5 (a prime no.)

(ii) n²-2n+2=1

From here n=1 & n⁴+4=5 (a prime no.)

(iii) n²+2n+2=-1

From here n is not an integer

(iv) n²-2n+2=-1

From here n is not an integer

So n=1,-1 (2 values of n)

Let $n$ be an arbitrary integer. We consider $n^{4}+4$ mod 5.

Fermat's Little Theorem tells us that $n^{4}\equiv 1\mod{5}$ .

Therefore we see $n^{4}+4\equiv 1+4\equiv 0\mod{5}$ , so that 5 will divide $n^{4}+4$ for all $n$ , since it was arbitrary.

The only integers that will make the equation be prime is $\pm 1$ , since $(\pm 1)^{4}+4=5$ , and for all other values of integers, then the equation will be bigger than 5, thus having 5 as a divisor (by the above argument), so it cannot be prime.