Why n^4 + 1 is not prime for any integer n?

Question

prologue
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I assigned to my high school students (in the USA they are called K9, AFAIK) to prove that for any integer n > 1, (n^3 + 1) is a composite number. As you can easily verify it is true because
n^3 + 1 = (n + 1)(n^2 - n + 1)
thus for any n > 1, (n^3 + 1) is the product of two integer greater than one, then it is composite.

trying to apply the converse
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The polynomial
P(n) = 16n^4 + 1 has no factors over ℚ
nevertheless it fails to be prime for n = 4
4097 = 17·241
for n = 5
10001 = 73·137
and many others

what confuses me
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Over the reals
P(n) = (4 n^2 + 2 n√2 + 1) (4 n^2 - 2 n√2 + 1)
plugging 4
P(4) = (65 + 8√2) (65 - 8√2)
gives no hints to the actual factorization

Over the complex field
P(n) = 16 (n + √2/4 + √2 i/4) (n + √2/4 - √2 i/4) (n - √2/4 + √2 i/4) (n - √2/4 - √2 i/4)
it is even less evident the reason why P(4) is composite

the question
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Why the factorization of the polynomial over Z proves that the numbers generated are all composite, while the irreducible polynomials do not generate prime numbers for any value of the variable?

Any hint/suggestion/clue would be great!

thank you in advance

Answer 1

This is a statement, proven by Christian Goldbach: no non-constant polynomial with integer coefficients exists, such that all its values for natural argument are prime.

Proof: let f(x) is a polynomial with integer coefficients, deg(f) = n, and let

f(x₀) = p for some x₀ - natural and p - prime. According the Taylor's Formula

f(x₀ + pt) = f(x₀) + pt * f ' (x₀) + (pt)² * f " (x₀)/2! + . . . + (pt)ⁿ * f⁽ ⁿ ⁾(x₀)/n!

f(x₀ + pt) is divisible by p because the right hand side is divisible by p (t - integer).

If all values of f for integer t are prime, then the above implies f(x₀ + pt) = p for all integer t, but when t → ∞ this implies that f(x) = const = p - contradiction.



Edit: Much harder problem is the distribution of primes among the values of a polynomial. For 1st degree polynomials the main result is the famous Dirichlet's Theorem:

http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions

For polynomials of higher degree almost nothing is known: for example are there finite or infinite prime values among the values of x² + 1 is an open problem (see 'Generalizations' in the article). I am afraid that your polynomial x⁴ + 1 will be a very hard nut to crack.

Answer 2

In keeping with your other response for n^4 + 1 not prime, when n = 3, (3)^4 + 1 = 82 which is obviously not prime. Looking at an integer domain & range gave this result. Obviously it is a continuous function over a domain from (staying with prime numbers), it is a matter of finding an integer mapping from the set of the domain to the range, if one exists.



This one turned out to be not so difficult.









** You teach this is USA high schools??

Answer 3

Why n^4 + 1 is not prime for any integer n?

That is false statement as n = 4 then this thing

collapses.