Let n be a positive integer. Show that (n^4) + 4 is prime iff n = 1. ?

Answer 1

If n = 1, the result is true.
We have to show that if n > 1, n^4 +4 is composite.
Write it as n^4 + 4n^2 + 4 - 4n^2 = (n^2 +2)^2 -(2n)^2 =
= (n^2 + 2n +2)(n^2-2n+2),
so n^4 + 4 is not prime if n > 1, since then both
factors can be shown to be greater than 1.

Answer 2

I think the only way to prove this is by cases... One direction is trivial, so I assume you can do that on your own... Here's a proof for some of the rest of it.
suppose n>1. (For the sake of deriving a contradiction.)
For ease i'm going to let (n^4)+4=f(n)
Now your goal is to show that in all possible cases f(n) is composite.
For all even n, f(n) is divisible by two, so f(n) is composite.
For all n such that n ends in the digit 1, 3, 7, or 9 f(n) is divisble by 5, so f(n) is composite. (Except in the case n=1, because 5 is prime)
That takes care of all the cases except the cases in which the last digit in n is 5. I can't figure that one out... hopefully someone smarter than me can help you there.

Nevermind... Steiner's proof is far more elegant than mine. (And covers all cases)

Answer 3

Don't u have the value of A,
the value of the above sum will be prime when A will be a proportionate number if n=1