7
$\begingroup$

Determine for what numbers $n$ the number $n^4 + 4$ is a composite number.

Sorry about my English. I found $n^4 + 4 = (n^2 + 2n +2)(n^2 -2n + 2)$, but i don't know what to do from here.

  • 0
    You have done the hard bit (the factorisation) - now you just need to make the conclusion that if the 2 factors are greater than 1, then the number must be composite.2012-06-10
  • 0
    (Anecdotal: This is actually Sophie Germain's identity with $b=1$: http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity.)2012-06-10

2 Answers 2

9

You did the non-obvious part, and are now essentially finished! Determine the $n$ for which one of your terms could be $\pm 1$. Not many! (And neither can be $-1$.) For all other $n$, you will have a non-trivial factorization. It may be useful to note that $n^2-2n+2=(n-1)^2+1$, with something similar for the other one.

Added: It turns out that in effect the OP wondered whether $n^4+4$ can be a prime power. Except when $n=0$, it cannot.

If a prime $p \gt 2$ divides both $n^2-2n+2$ and $n^2+2n+2$, then $p$ divides $n$, but then $p$ divides $2$, contradiction. So the only possibilities are $n^2+2n+2$, $n^2-2n+2$ both a non-trivial power of $2$, Then $n$ has to be even. It follows that each of $n^2-2n+2$ and $n^2+2n+2$ is congruent to $2$ modulo $4$, so if each is a power of $2$, each must be $2$, giving $n=0$.

  • 0
    I don't understand one thing. It might happen that the first term being, for example, a prime and the second term that is a exponent of this prime?2012-06-10
  • 0
    @jonjones A composite number $n$ is a number with two factors different from $\pm 1$ and $\pm n$. But the factors don't even have to be different. $2 \times 2 = 4$ is composite, because it has two factors that aren't $\pm 1$ or $\pm 4$, even though these two factors are equal. And in the case you mention, if $p$ is prime, $p \times p^k = p^{k+1}$ is also prime, even though all its prime factors are equal to $p$. All that is needed to prove that $n$ is composite is to exhibit one or two factors of $n$ different from $\pm 1$ and $\pm n$. (Note that if you find one such factor $m$, then ...2012-06-10
  • 0
    hmmmm sorry. I though that the definition was different. Really sorry and thanks.2012-06-10
  • 0
    @jonjones ... you have two: $\frac{n}{m}$ is another one.) No problem.2012-06-10
  • 0
    @jonjones: We can ask whether $n^4+4$ can be a non-trivial **power** of a prime. Interesting variant! It can be, take $n=0$. But that's it. If a prime $p \gt 2$ divides both $n^2-2n+2$ and $n^2+2n+2$,then $p$ divides $n$, but then $p$ divides $2$, contradiction. So the only possibilities are $n^2+2n+2$, $n^2-2n+2$ both a non-trivial power of $2$, Then $n$ has to be even. This quickly leads to $n=0$.2012-06-10
0

If $n > 1$ then $n^4+4$ is composite

  • 2
    Please elaborate your answer by supplying some motivation. Additionally, please see the help on how to type formulas.2013-03-07
  • 0
    Please explain why.2013-03-07