Prove: For integers n>1, n⁴ + 4 is not a prime.

If you worked on this for any length of time without success, you'll kick yourself now: n⁴ + 4 can be factored.  Here it is:

n⁴ + 4
n⁴ + 4n² + 4 - 4n²
( n⁴ + 4n² + 4)  - 4n²
((n² + 2) + 2n)((n² + 2) - 2n)

And since n>1 you can show that both factors are larger than one.  You're done.

A similar problem was suggested to me, to prove that n⁴-20n²+4 is not prime.  This one had me stumped for a while because I was trying this:

n⁴ - 20n² + 4
n⁴ + 4n² + 4 - 24n²
( n⁴ + 4n² + 4)  - 24n²
((n² + 2) + sqrt(24)(n))((n² + 2) - sqrt(24(n))

But those factors, while real, are not integers.  So it doesn't help in the effort to show compositude.  Sorry, I can't help making up new words.  Here's the real answer:

n⁴ - 20n² + 4
n⁴ - 4n² + 4 - 16n²
( n⁴ - 4n² + 4)  - 16n²
((n² - 2) + 4n)((n² - 2) - 4n)

Here, the first factor is greater than one if n>0 -- that's the point where the n² and n terms overpower the constant. The second factor is greater than one if n>4, where the n² term overpowers the n term.  So the polynomial is composite for all n>4.  It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60.

Related pages in this website

The webmaster and author of the Math Help site is Graeme McRae.
     [home]  [email]  [search]  [Links to Math Sites]  [Whiteboard]