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The obvious choice seemed at first to be $\mathbb{Z}[\root 4 \of 4]$. But since I know next to nothing about quartic fields, I thought to look in the quadratics.

For the first few such primes in $\mathbb{Z}[\sqrt{-2}]$ I was doing pretty well, e.g., $3^4 + 4^2 = 97 = (5 - 6 \sqrt{-2})(5 + 6 \sqrt{-2})$. I thought I hit a wall with $2417$ but then I found a mistake in my calculations which led me to $(45 - 14 \sqrt{-2})(45 + 14 \sqrt{-2})$.

Of course I could find a thousand primes of this form that split in this way in $\mathbb{Z}[\sqrt{-2}]$, but that doesn't rule out the possibility of a single inert prime just beyond the reach of my calculations.

Is $\mathbb{Z}[\sqrt{-2}]$ the ring I'm looking for? And if not, what is?

James47
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    I'm not sure about $\mathbb{Z}[\sqrt{-2}]$, but in $\mathbb{Z}[i]$ we get $a^4 + 4^b = (a^2 + i\cdot 2^b)(a^2 - i\cdot 2^b)$. – Marc Jan 25 '16 at 23:09
  • Be careful in your first line - $\sqrt[4]4 =\sqrt2$! – Mathmo123 Jan 25 '16 at 23:35
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    I think this other question should be listed as a related question http://math.stackexchange.com/questions/1622295/proof-of-a-prime-number/1626982#1626982 since at the very least it can be shown to have motivated this one. – Robert Soupe Jan 26 '16 at 02:25
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    @Mathmo123 Do you mean that $\root 4 \of 4 \approx 1.25381548$? Wink, wink. – Robert Soupe Jan 26 '16 at 02:29
  • @RobertSoupe yes linking that question would've made finding the factorisations much easier. – Mathmo123 Jan 26 '16 at 07:56
  • @Robert Its marked as a linked question. Did the system do that automatically or did someone do something I neglected to do? – James47 Jan 29 '16 at 22:25
  • @James I'm not sure, maybe the system did it automatically when I posted my previous comment. – Robert Soupe Jan 30 '16 at 02:44

2 Answers2

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Suppose $p = a^4+4^b$ for some non-negative integer $b$.

If $b\equiv 0\pmod 2$, then we can write $$\begin{align}p&=(a^2+2^{b}i)(a^2-2^{b}i)\\&=(a^2+2^{b/2}\sqrt{2}a+2^b)(a^2-2^{b/2}\sqrt 2a+2^b)\\&=(a^2+2^{b/2}\sqrt{-2}a-2^b)(a^2-2^{b/2}\sqrt {-2}a-2^b)\end{align}$$

so $p$ splits in any of $\mathbb Z[i],\mathbb Z[\sqrt2],\mathbb Z[\sqrt{-2}]$.

(We should probably assume that $p$ is an odd prime, since $2 = 1^4+4^0$ is ramified in all these rings.)

If $b\equiv 1\pmod 2$, then letting $b'=\frac{b+1}2$ we have $$p=(a^2+2^{b'}a+2^{b})(a^2-2^{b'}a+2^{b}),$$ so assuming, wlog, that $a\ge 0$, we see that $p$ can be prime only if $$a^2-2^{b'}a+2^b = \pm1$$which, solving for $a$ as a quadratic, can occur only when $$2^{2b'}-4(2^b\mp1)=4(2^{b-1}-2^b\pm1)=4(\pm1-2^{b-1})$$is a perfect square - i.e. $a=b=1$ and $p=5$.

Now $5=(1+2i)(1-2i)$ splits in $\mathbb Z[i]$, but not in $\mathbb Z[\sqrt{\pm 2}]$ since $X^2\pm 2$ is irreducible modulo $5$.

It follows that $\mathbb Z[i]$ is the ring you're after.

Mathmo123
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I think Mathmo123 already gave the Best Answer (and the one which deserves the bounty), and I understand your trepidation at dealing with quartic domains. But I also think you were a bit too hasty to drop the quartic domains line of inquiry, even though it still leads to a quadratic domain as the answer.

For now I'm going to ignore the special case $a = b = 1$ to concentrate on $b$ an even nonnegative integer (and then of course $a$ has to be odd).

Suppose there is a number $\theta$, possibly complex (I don't know at this point), such that $\theta^4 - 1 = 0$, so $\mathbb{Z}[\theta]$ is almost certainly a quartic domain. Then $$(a^2 - 2^{\frac{b}{2}})(a^2 + 2^{\frac{b}{2}}) = a^4 - 4^b \theta^2.$$

Clearly my thinking is still dominated by quadratic domains. But if it were the case that $\theta^2 = -1$, we would be done, as the right hand side of that equation would boil down to $a^4 + 4^b$. In fact, the only way to have $\theta^2 = -1$ and $\theta^4 = 1$ is for $\theta$ to be $i$ or $-i$. Then it turns out that $\mathbb{Z}[\theta] = \mathbb{Z}[i]$, a quadratic domain after all. And it works for the special case, too, as we have $1^4 + 4^1 = 5$ and $5 = (2 - i)(2 + i)$.

What about $\theta^4 + 1 = 0$? Then $\theta = \pm \sqrt{i}$. That's still a bit beyond my ability at this point.

Mr. Brooks
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