MHB Proving the Compositeness of n^4+1: A Scientific Approach

[evinda]

Hello! (Smile)

I am looking at the following exercise:

If $n>1$,show that $n^4+1$ is composite.

According to my notes,it is like that:

$$n^4+4=(n^2)^2+2^2+2 \cdot 2 \cdot n^2-2 \cdot 2 \cdot n^2=(n^2+2)^2-4n^2=(n^2+2-2n)(n^2+2+2n)$$

To check if it is composite,we have to check if $n^2+2-2n>1$.

The condition is satisfied,so it is composite.

But...is there also an other way we could show that $n^4+1$ is composite ? (Thinking) (Thinking)

 

[magneto1]

There might be a typo: $x^4 + 1$ is actually prime when $x = 2$, do you mean $x^4 + 4$ as your example show?

 

[magneto1]

For the benefits of others, this type of factoring actually has a name: Sophie Germain Identity.

I suspect that is likely the best you can do for this problem. Otherwise, you can consider the case of $n$ even, which we can see $n^4 + 4$ is even for $n > 1$, and therefore divisible by $2$ (actually you can show it is divisible by $4$), so it is composite. For $n$ odd, the pattern is a bit more difficult to find.

$n = 3$, $n^4 + 4 = 5 \cdot 17$.

$n = 5$, $n^4 + 4 = 17 \cdot 37$.

$n = 5$, $n^4 + 4 = 5 \cdot 13 \cdot 37$.

$n = 9$, $n^4 + 4 = 5 \cdot 13 \cdot 101$.

 

[I like Serena]

But...is there also an other way we could show that $n^4+1$ is composite ? (Thinking) (Thinking)

Hmm...

To show it is composite, I think we really need to factorize it... (Thinking)

Or else we need to assume it is prime and show that cannot be true... by coming up with a factorization... (Doh)

 

[evinda]

Hmm...

To show it is composite, I think we really need to factorize it... (Thinking)

Or else we need to assume it is prime and show that cannot be true... by coming up with a factorization... (Doh)

A ok..I understand! Thank you very much! (Mmm)

 

[evinda]

There might be a typo: $x^4 + 1$ is actually prime when $x = 2$, do you mean $x^4 + 4$ as your example show?

In my notes, there is the exercise that I have to show that $n^4+1$ is composite.. Maybe it is a typo,I don't know.. (Thinking) (Worried) It is actually a prime for $x = 2$.. (Thinking)

 

[I like Serena]

In my notes, there is the exercise that I have to show that $n^4+1$ is composite.. Maybe it is a typo,I don't know.. (Thinking) (Worried) It is actually a prime for $x = 2$.. (Thinking)

Well... I know!
$n^4+1$ is actually prime for $x = 2$! (Evilgrin)

Actually, a hand written 1 can be rather close to a hand written 4... (Thinking)

 

[evinda]

Well... I know!
$n^4+1$ is actually prime for $x = 2$! (Evilgrin)

Actually, a hand written 1 can be rather close to a hand written 4... (Thinking)

Yes,it is possible that it was meant the number $4$,and not the number $1$.. (Blush)

But, isn't it always $n^2+2-2n>1, \forall n \in \mathbb{Z}$ ??

 

[I like Serena]

But, isn't it always $n^2+2-2n>1, \forall n \in \mathbb{Z}$ ??

Not always!
Only if $n>1$.

 

[evinda]

Not always!
Only if $n>1$.

And..it our case it is $n>1$.So,why can it happen that $n^4+1$ is not composite,for a $n>1$ ? (Thinking) (Thinking)

 

[Nono713]

I don't see what the problem is. Your notes (as given in your original post) clearly show that $n^4 + 4$ is composite for all $n > 1$, which is true. But you cannot do "the same thing" for $n^4 + 1$, because the claim is simply false, as noted above $2^4 + 1 = 17$ which is prime, $4^4 + 1 = 257$ which is prime, $6^4 + 1 = 1297$ which is prime, etc, etc. The factorization ILS is working with is for $n^4 + 4$, not $n^4 + 1$.

 

[evinda]

I don't see what the problem is. Your notes (as given in your original post) clearly show that $n^4 + 4$ is composite for all $n > 1$, which is true. But you cannot do "the same thing" for $n^4 + 1$, because the claim is simply false, as noted above $2^4 + 1 = 17$ which is prime, $4^4 + 1 = 257$ which is prime, $6^4 + 1 = 1297$ which is prime, etc, etc. The factorization ILS is working with is for $n^4 + 4$, not $n^4 + 1$.

Ok,I understand...thanks a lot! (Smile)