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There is a famous problem that says that $a^4+4, a\in\Bbb N$ cannot be prime except than $a=1$. The solution based on factoring:

$$a^4+4=(a^2+2)^2-4a^2=(a^2+2-2a)(a^2+2+2a)$$

Usually, a problem in mathematics has more than one solution method. But for this question, I couldn't find any way other than factoring. As far as I can tell, it's nowhere to be found. So, I have a natural question. Is there any known way around this problem other than factoring?

This way can be induction. It can be modular arithmetic or some techniques of number theory.

Clearly $a$ can not be even, and if $a=5k+n, 1\leq n\leq 4$, then $5\mid a^4+4$. But this doesn't work when $a\not\mid 5$.

But, I haven't seen any helpful hints that these work yet.

Bill Dubuque
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User
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  • You can ask this also for $a^4+4^n$ more generally, see here. Often a problem in mathematics has more or less only one solution. – Dietrich Burde Nov 30 '22 at 15:37
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    @DietrichBurde Yes, there is no way except than factoring. – User Nov 30 '22 at 15:39
  • 1, 2. – Jyrki Lahtonen Nov 30 '22 at 16:08
  • Since composite means factorable it seems a proof, at best, is only going to avoid its factorization superficially and all proofs will some at least obliquely refer to its having a factorization. Or so I feel is likely to happen. – fleablood Nov 30 '22 at 16:50
  • As I emphasize in this answer, a simple natural way to discover such factorizations is simply to note that when $,\color{#c00}{2ab=c^2},$ is a square then completing the square yields a difference of squares

    $$\large\begin{align}a^2!+b^2 &= (a!+!b)^2!-\overbrace{\color{#c00}{2ab}}^{\textstyle \color{#c00}{c^2}}\[.2em] &= (a!+!b!-!c)(a!+!b!+!c)\end{align}\qquad$$

    See the linked post for many examples.

     – Bill Dubuque Nov 30 '22 at 17:01
  • Generally induction is not a feasible method of proving (or discovering) such - esp. for more complex generalizations such as Aurifeuillian factorizations of cyclotomics, e.g. below

    $$\dfrac{x^{12} + 6^6}{x^4 + 36} = (x^4 + 6x^3 + 18x^2 + 36x + 36);(x^4 - 6x^3 + 18x^2 - 36x + 36)\qquad$$

     – Bill Dubuque Nov 30 '22 at 17:01
  • @BillDubuque As far as I understand it, even if the induction method or the method proof by contradiction works, these cannot escape factoring, right? – User Nov 30 '22 at 17:12
  • It is generally true that in number theory we can prove many results about "numbers" as specializations of "functions" (polynomials) because polynomials enjoy much richer structure (e.g. we can differentiate them, leading to (squarefree) factorization algorithms, etc). A good example is the two-line proof of FLT for polynomials. So your sought solution method is the opposite of the natural way to simplify such problems. Follow the link for further discussion of such. – Bill Dubuque Nov 30 '22 at 17:18
  • See also this answer for literature that relates primality (compositeness) of integers vs. polynomials, e.g. Bouniakowski's conjecture. – Bill Dubuque Nov 30 '22 at 17:24
  • @fleablood That "composite means factorable" doesn't mean a proof of compositeness has to refer even indirectly to the number having a factorization. Primes have properties, so if a number doesn't have that property, then it's composite. For example, if $n$ is prime then $a^{n-1} \equiv 1 \bmod n$ for all $a$ from $1$ to $n-1$, so if $a^{n-1} \not\equiv 1 \bmod n$ for some $a$ in that range, then $n$ is composite but we don't learn a factorization. The number $F_{14} = 2^{2^{14}}+1$ was proved composite in $1961$ but a nontrivial factor was first found almost $50$ years later, in $2010$. – KCd Nov 30 '22 at 17:54

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If $a$ is divisible by $2$ then $a^4 + 4$ is a multiple of $4$.

If $a$ is not divisible by $5$ then $a^4 + 4$ is a multiple of 5 by modular arithmetic: $a^4 \equiv 1 \bmod 5$, so $a^4 + 4 \equiv 0 \bmod 5$.

If $a$ is an odd multiple of $5$ then I think this is a more subtle issue if you don't "see" that algebraic factorization of $a^4+4$ because there is no universal common prime factor, and in fact the two factors from the algebraic factorization can often both be prime:

$5^4 + 4 = 17 \cdot 37$, $15^4 + 4 = 197 \cdot 257$, $25^4 + 4 = 577 \cdot 677$, $55^4 + 4 = 2917 \cdot 3137$, $125^4 + 4 = 15377 \cdot 15877$

The difference of these prime factors has an easily spotted pattern: it is $4a$ when $a$ is an odd multiple of $5$, and armed with this information you might be led to discover the algebraic factorization of $a^4 + 4$ for all $a$.

Remark. As far as I am aware, the only reason $a^4 + 4$ is famous in math is due to its "unexpected" algebraic factorization. So that factorization is a good thing and not something to avoid knowing.

KCd
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  • Well, do you have a proof that induction can never work? – User Nov 30 '22 at 15:50
  • Proving compositeness by induction sounds like a strange idea. Where have you ever seen that done? – KCd Nov 30 '22 at 15:54
  • I'm not claiming that factoring isn't a good thing. The problem here is a question of wondering if there is "different" way to solve this problem. – User Nov 30 '22 at 15:55
  • The factorization is attributed to Sophie Germain – J. W. Tanner Nov 30 '22 at 16:02
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    Bunyakovsky's conjecture implies that if a polynomial $f(x)$ with integral coefficients and leading coefficient $1$ has a pair of relatively prime values (like $x^4 + 4$, whose values at $0$ and $1$ are $4$ and $5$), then $f(x)$ will have infinitely many prime values if and only if $f(x)$ is irreducible in $\mathbf Z[x]$. So if you don't see a prime value for $f(x)$ at positive integers other than at $x = 1$, then you should believe $f(x)$ is in fact reducible in $\mathbf Z[x]$, and that is enough motivation to search for a nontrivial factorization of $f(x)$. – KCd Nov 30 '22 at 16:03
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    If you want to motivate others to find a way to prove $a^4 + 4$ is always composite that doesn't rely on its algebraic factorization, then can you offer first a proof that $a^2 - 1$ is always composite for $a > 2$ that does not rely on its factorization as $(a+1)(a-1)$? Note that when $a$ is the number between two twin primes, $a^2-1$ is going to be a product of two different primes (for example, $11$ and $13$ are a pair of twin primes and $12^2 - 1 = 143 = 11 \cdot 13$), and we expect there are infinitely many twin primes. – KCd Nov 30 '22 at 16:07