Some composite numbers

Why is the number n^{4} + 4 composite for integers n ≥ 2? One needs n ≥ 2 as if n = 1 then 1^{4} + 4 is prime. The best way of proving a number compoisite is to factorize it so let’s try to factorize n^{4} + 4. More generally let’s try to factorize x^{4} + 4. The standard way of doing this is to solve x^{4} + 4 = 0. Obviously this has no real roots, but we’re not afraid of complex numbers are we? This equation says that x^{4} = −4 so let’s write x = re^{iθ} with r > 0. Then

x^{4} = r^{4}e^{i4θ} = −4 = 4e^{iπ}.

Thus r = ^{4} \sqrt{4} = \sqrt{2} and 4θ = π + 2kπ for some integer k. We need four solutions, so take k going from 0 to 3 to get θ = π/4, 3π/4, 5π/4 or 7π/4. The roots of the equation are

x = \sqrt{2}e^{iπ/4} = \sqrt{2}(cos π/4 + i sin π/4) = \sqrt{2} (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) = 1 + i,

x = \sqrt{2}e^{i3π/4} = \sqrt{2}(cos 3π/4 + i sin 3π/4) = \sqrt{2} (−\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) = −1 + i,

x = \sqrt{2}e^{i5π/4} = \sqrt{2}(cos 5π/4 + i sin 5π/4) = \sqrt{2} (−\frac{1}{\sqrt{2}} − i\frac{1}{\sqrt{2}}) = −1 − i

x = \sqrt{2}e^{i7π/4} = \sqrt{2}(cos 7π/4 + i sin 7π/4) = \sqrt{2} (\frac{1}{\sqrt{2}} − i\frac{1}{\sqrt{2}}) = 1 − i.

Thus

x^{4} + 4 = (x − 1 − i)(x + 1 − i)(x + 1 + i)(x − 1 + i).  (1)

This is very impressive but putting x = n in this will factorize n^{4} + 4 into complex numbers which is not what we want. So in (1) we pair off conjugate terms. Note that

(x − 1 − i)(x − 1 + i) = (x − 1)^{2} − i^{2} = x^{2} − 2x + 2

and

(x + 1 − i)(x + 1 + i) = (x + 1)^{2} − i^{2} = x^{2} + 2x + 2.

Thus

x^{4} + 4 = (x^{2} − 2x + 2)(x^{2} + 2x + 2)  (2)

and so

n^{4} + 4 = (n^{2} − 2n + 2)(n^{2} + 2n + 2)

So n^{4}+4 factors! But we must be careful; are these factors positive, and is either equal to 1? But if n ≥ 2, n^{2}−2n+2 = (n−1)^{2}+1 ≥ 1+1 = 2 and n^{2}+2n+2 = (n+1)^{2}+1 ≥ 9+1 = 10. Indeed n^{4} + 4 is composite for n ≥ 1.

Now let’s look at n^{4} + 4^{n}. Again we need n ≥ 2. If n is even so is n^{4} + 4^{n} so we may assume that n is odd. Write n = 2k+1 so that n^{4}+4^{n} = n^{4}+4^{2k+1} = n^{4}+4·4^{2k} = n^{4}+4·2^{4k}.

This has the form x^{4} + 4y^{4} with x = n and y = 2^{k}. But from (2) we can deduce a factorization of x^{4} + 4y^{4}; in detail,

x^{4}+4y^{4} = y^{4} (\frac{x^{4}}{y^{4}} + 4) = y^{4} (\frac{x^{2}}{y^{2}} − \frac{2x}{y} + 2) (\frac{x^{2}}{y^{2}} + \frac{2x}{y} + 2) = (x^{2}−2xy+2y^{2})(x^{2}+2xy+2y^{2}).

(3)

Consequently

n^{4} + 4^{n} = (n^{2} − 2^{k+1}n + 2^{2k+1})(n^{2} + 2^{k+1}n + 2^{2k+1}).

Again we need to check that these factors are each at least 2. The former factor is smaller and

n^{2} − 2^{k+1}n + 2^{2k+1} = (n − 2^{k})^{2} + 2^{2k} ≥ 2^{2k} ≥ 2^{2} = 4

as k ≥ 1. Thus n^{4} + 4^{n} is indeed composite for n ≥ 1.

RJC 4/3/97