Proving Infinitely Many Natural Numbers: Larson 4.1.6

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[SOLVED] Larson 4.1.6

Homework Statement


Prove that there are infinitely many natural numbers a with the following property: The number n^4+a is not prime for any number n.

Homework Equations


The Attempt at a Solution


I cannot even think of one such natural number a. :(
I need to find some way to factor this after we put some restrictions on a. That is we need to express a in a special form that makes this factorable. If a is equal to b^4, it is not necessarily factorable. In fact, I don't know of any power of b that will make it factorable. a cannot be a function of n. I really don't know what to do.
 
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a=4
a=(multiple of 5)-1
hence infinite
i guess
correct me if iam wrong
have you any idea of fermat theorm
n^5-n is divisible by 5
can be prooved ,it is a simpler form of fermat theorm
n(n^4-1) certainly n^4 -1 is divisible by 5
add any multiple of 5 to it
you get
 
Fermat's Little Theorem says that if p is a prime number that does not divide an integer n, then n^{p-1} \equiv 1 \mod p.

Therefore that will only apply when n is not divisible by 5. We need a proof for all n in N.
 
Maybe you can use Sophie Germain's identity:
a^4 + 4 b^4 = (a^2 + 2 b^2 + 2 a b) (a^2 + 2 b^2 - 2 a b).
 
durt said:
Maybe you can use Sophie Germain's identity:
a^4 + 4 b^4 = (a^2 + 2 b^2 + 2 a b) (a^2 + 2 b^2 - 2 a b).

Wow. Thanks. I'm glad I posted this question because I never would have thought of that.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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