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Consider: $X^5=Y^2+4$ $X,Y \in \bf Z$.

If $11 \not | X \rightarrow X^{10}-1\equiv 0 \pmod {11}\rightarrow (X^5-1)(X^5+1)\equiv 0 \pmod {11} $.

From here we do a remainder table for all numbers and see that there's $Y: Y^2+4$ that can be congruent to $1\bmod 11$.

What's a more elegant and less magical way to prove this?

YoTengoUnLCD
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  • Less magical, but much longer, is factorization of $Y^2+4$ as $(Y+2i)(Y-2i)$ in the Gaussian integers. Then argue that the factors have to be both fifth powers, and calculate. – André Nicolas Oct 24 '15 at 04:39
  • @AndréNicolas First you'd have to establish that $\gcd(Y+2i, Y-2i)=1$. If $d\mid Y+2i, Y-2i$, then $d\mid 2Y, 4i$, so $N(d)\mid 4Y^2, 16$. If $Y$ is even, then mod $16$ gives a contradiction, so $Y$ is odd. But then we only know $N(d)\mid 4$, so the coprimeness is not obvious. – user236182 Oct 24 '15 at 05:20
  • They are not necessarily relatively prime, there is a somewhat different argument when they are not, we have to look at $8Z^5=W^2+1$. For fun or out of masochism I did the calculations. I was going to write it up, but too much typing. – André Nicolas Oct 24 '15 at 05:40
  • @AndréNicolas $8Z^5=W^2+1$ is impossible mod $4$. Clearly $Y$ is odd. – user236182 Oct 24 '15 at 06:03
  • Hadn't noticed, did that part in $\mathbb{Z}[i]$ too! – André Nicolas Oct 24 '15 at 06:12
  • With $Y$ odd the coprimality is clear, for any common (Gaussian) prime divisor divides $2$. But then since it divides $Y-2i$, it divides $Y$, which is impossible for odd $Y$. – André Nicolas Oct 24 '15 at 06:20
  • @AndréNicolas oh sorry, indeed I got that $N(d)\mid 4$, and since $d\mid Y+2i\implies N(d)\mid Y^2+4=X^5$ and $X$ is odd, we get $N(d)=1$, so $\gcd(Y+2i,Y-2i)=1$. – user236182 Oct 24 '15 at 06:28
  • @AndréNicolas I've written an answer using $\Bbb Z[i]$. – user236182 Oct 24 '15 at 06:49
  • Your solution uses that $X^5\equiv {-1,0,1}\pmod{11}$ and $Y^2+4$ is never in ${-1,0,1}$ mod $11$ (your description is unclear). – user236182 Oct 24 '15 at 07:06

1 Answers1

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If $Y$ is even, we get a contradiction mod $16$. So $Y$ is odd. We'll be working in $\Bbb Z[\sqrt{-1}]$, i.e. $\Bbb Z[i]$, i.e. the Gaussian integers.

$X^5=(Y+2i)(Y-2i)$. First we'll prove $Y+2i, Y-2i$ are coprime (i.e. relatively prime). Suppose $d\mid Y+2i, Y-2i$. Then $d\mid (Y+2i)-(Y-2i)=4i$, so $N(d)\mid 16=2^4$, where $N(d)$ denotes the norm of $d$ (i.e. if $d=x+yi$, then $N(d)=x^2+y^2$).

But also $d\mid Y+2i\implies N(d)\mid Y^2+4=X^5$, which is odd. We conclude $N(d)=1$. If $d=x+yi$, then $N(d)=x^2+y^2=1$, so $\{x,y\}=\{\pm 1, 0\}$, so $d$ is a unit. So $Y+2i, Y-2i$ are coprime.

Since $\Bbb Z[i]$ is a Unique Factorization Domain (i.e. UFD), we get $Y+2i=u(a+bi)^5$ for some $a,b\in\Bbb Z$ and a unit $u\in\{\pm i, \pm 1\}$; but since $\pm i, \pm 1$ are fifth powers, we can just let $Y+2i=(a+bi)^5$.

Therefore $5a^4b-10a^2b^3+b^5=2$, so $b\mid 2$. Let $a^2=t$. Testing $b\in\{-2,-1,1,2\}$ and solving quadratic equations in $t$ gives no solutions.

Remark: $\Bbb Z[\sqrt{-2}]$ is also a UFD. However, if $n\ge 3$, then $\Bbb Z[\sqrt{-n}]$ is not a UFD (see here). Due to the nature of units of $\Bbb Z[\sqrt{n}]$ when $n\ge 2$ (infinitely many units described by $x^2-ny^2=1$, related to Pell's equations), such methods couldn't be easily applicable (even though some of them can be UFDs). So only use this method for $\Bbb Z[\sqrt{-2}], \Bbb Z[\sqrt{-1}]$.

For more practice, see this paper (about Mordell's equations), where elementary similar examples are given.

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