We shall prove that $x$ is odd. Indeed, suppose that $x$ is even. Then $y$ is also even, so it's divisible by $16$. It means that $x^2 \equiv 12 \mod{16}$ but that's impossible (it can be checked by hand very easily). Now, we see that the equation is equivalent to the following: $(x+2i)(x-2i) = y^5$ Now let $d=(x+2i, x-2i)$ It follows that $d|4i$ But because $x$ is odd we must have $d|i$. It implies that $x+2i$ and $x-2i$ are coprime, so they have to be the fifth powers. $x+2i = (a+bi)^5$ and $x-2i = (a-bi)^5$ (this is because $x+2i$ and $x-2i$ are conjugates) Now we substract the second equation from the first and, after doing some boring, bot fortunately not too long, computations, we obtain a equation which has no solutions (after comparing the real and imaginary parts or so...).

I remmeber (mod 11) can solved the problem. But Which book introduced the knowledge of $Z[i]$?I want to learn it by myself.it seems very useful. ??:

Hawk Tiger wrote: I remmeber (mod 11) can solved the problem. But Which book introduced the knowledge of $Z[i]$?I want to learn it by myself.it seems very useful. ??: I strongly recommend the book Theory of Numbers: A Text and Source Book of Problems written by Andrew Adler and John E. Cloury.

my solution is more simple. first, is easy to see that x is odd. now, we see that $x^2 + 36 = (y+2)(y^4-2y^3+4y^2-8y+16)$ and that, if p is a prime of form 4k+3, and p|a+b, we have that p|a and p|b! (the two factors that compose the rhs, say A and B, have that 3 don't divide A and B at the same time. so, because A and B are of form 4k+3, one of this have one factor p of this form, different of 3, that divide x+36)

I think it's more usefull to directly learn superior algebra, at least the standard stuff on rings like to be factorial and on fields like Galois stuff. Then you can take a look at J. Neukirch: "Algebraic Number Theory" (best book ever ) and Ram Murty: "Problems in algebraic number theory" (being on a lower level and having lots of good problems, a good amount of them is here on the site).

Peter wrote: Prove that there are no integers $x$ and $y$ satisfying $x^{2}=y^{5}-4$. Here is a standard proof : Actually ; the initial equation is equivalent to : $ x^2 + 4 = y^5 \ \ (*) \implies y^5 + 2^5 = x^2 + 36$ $\implies x^2 + 6^2 = (y+2) \cdot (y^4 - 2y^3 + 4y^2 - 8y + 16 ) $ Consider $ b = y^4 - 2y^3 + 4y^2 - 8y + 16 $ ; we have : $ b \equiv y^4 - 2y^3 = y^4 - 2y^3 + y^2 - y^2 = y^2 ( y-1)^2 - y^2 \ \ \mod 4 \ \ (**)$ From $(*)$ ; we can deduce that $2 \not | y$ since if $y$ is an even number ; we can have a contradiction as given in Tomicio's post ( $x$ modulo $8$). So we only need to condider the case : $y = 4k \pm 1$ Combine this with $(**)$ ; we always have : $ b \equiv -1 \ \ \mod 4 \ \ (1)$ So ; $b$ have at least one prime divisor $p$ of the form $4m +3$ $ \implies p | x^2 + 6^2 \implies p | x ; 6 \implies p = 3 ; x = 3t $ ( $2 \not | t$) If $ 3 | y+2 \implies y \equiv -2 \ \ \mod 3 \implies b \equiv 5 \cdot 2^4 ( \mod 3)$ The last congruence cannot hold since $ 3 | b$ Thus ; $3 \not | y+2 \implies 3^2 | b \implies b = 9k ; k \in \mathbb{N}$ $ \implies x^2 + 6^2 = 9(t^2 + 4) = (n+2) n = 9(n+2)k \implies t^2 +2^2= (n+2)k$ So ; It's very easy to see that $k$ doesn't have any prime divisor of the form $4k+3$ ( $2$ is not divisible by any odd prime ) $ \implies k \equiv 1 \ \ \mod 4 \implies b = 9k \equiv 1 \ \ \mod 4$ This congruence also cannot hold due to $(1)$ So ; we have done .

extending on the comment of mod 11. We are motived to use mod 11 because by fermats little theorem we can see that y^5 = $\pm$ 1 mod 11 which is very constrained which is good in this situation. y^5-4 = 6,8 mod 11 however a simple check shows that the squares mod 11 are 1,4,9,5,3. Thus there are no solutions.

(modulo11) it's have x^5 congruence -1,1 (mod 11) <Fermat's little theorem> and x^2 congruence 8,6 (mod 11) but, x^2 congruence 0,1,3,4,5,9 (mod 11) thus,no integers x and y satisfying x^2=y^5-4

It seems ---Fermat--- specializes in providing solutions which were already given in the thread (see the post just above his), but not notice the obvious flaw. The residues modulo $11$ for $y^5$ can be $-1, 0,1$, not just $-1,1$, as stated. Luckily, that means $y^5 - 4 \in \{6,7,8\}$ modulo $11$, none a square residue, so the method still works.

Let us show that there are even no rational solutions. My solution uses the theory of hyperelliptic curves and the 2-Selmer group from Michael Stoll's lecture notes (google for Skript-ArithHypCurves-pub-screen.pdf - all references refer to that PDF). Sorry, it is in German, I will translate it later. \begin{theorem} Sei $C/\mathbf{Q}$ die glatte projektive hyperelliptische Kurve vom Geschlecht $2$ zur affinen Gleichung $y^2 = x^5 - 4 =: f(x) \in \mathbf{Q}[x]$ mit dem Grad $5 = 2g+1$ ungerade. Wir zeigen, dass $C(\mathbf{Q}) = \{[0:1:0]\}$ nur aus dem Punkt im Unendlichen besteht, also $C_\mathrm{aff}(\mathbf{Q}) = \emptyset$, via $C(\mathbf{Q}) \hookrightarrow J(\mathbf{Q}) = 0$ mit $J/\mathbf{Q}$ die Jacobische von $C$, eine Abelsche Varietät der Dimension $g = 2$. \end{theorem} \begin{proof} Die Stellen schlechter Reduktion sind genau $2$ und $5$, die Diskriminante von $f$ ist $2^{16}\cdot5^5$. Weil $f$ über $\mathbf{Q}_2$ irreduzibel ist (Newton-Polygon hat genau ein Segment mit Steigung $\frac{v_2(-4)}{5} \not\in \mathbf{Z})$, ist $f$ auch über $\mathbf{Q}$ irreduzibel. Weil der Grad $5 = 2g+1$ von $f$ ungerade ist, ist $J(\mathbf{Q})[2] = (\mathbf{Z}/2)^{1-1} = 0$ (Lemma 5.2). Es ist sogar $J(\mathbf{Q})_{\mathrm{tors}} = 0$: $J$ hat wie $C$ gute Reduktion bei $3$ und $11$ und $J(\mathbf{F}_3) \cong \mathbf{Z}/10$ und $J(\mathbf{F}_{11}) \cong \mathbf{Z}/41$ haben keine gemeinsame nichttriviale Untergruppe (Theorem 4.21). Es ist $J(\mathbf{Q}_p)[2] = 0$ für $p = 2,3,5,7,11$, da $f$ über diesen $\mathbf{Q}_p$ irreduzibel bleibt. Daher ist $\mathrm{Sel}^2(J/\mathbf{Q}) = J(\mathbf{Q})[2] = 0$ (theory from Chapter 5). Aus der exakten Sequenz \[ 0 \to J(\mathbf{Q})/2 \to \mathrm{Sel}^2(J/\mathbf{Q}) \to \mathrm{Sha}(J/\mathbf{Q})[2] \to 0 \]folgt $J(\mathbf{Q}) = 0$ wegen $J(\mathbf{Q})_{\mathrm{tors}} = 0$ (und es folgt auch $\mathrm{Sha}(J/\mathbf{Q})[2^\infty] = 0$). Also ist der einzige Punkt von $C/\mathbf{Q}$ der Punkt $[0:1:0]$ im Unendlichen, also hat die ursprüngliche Gleichung keine rationale Lösung. \end{proof} \begin{remark} \begin{enumerate} \item Ganzzahlige Lösungen gehen leichter mit imaginär-quadratischen Zahlkörpern (see the posts above). \item Chabauty mit $r = 0 < 2 = g$ und $3 > 2r$ eine Stelle guter Reduktion gibt nur $|C(\mathbf{Q})| \leq |C(\mathbf{F}_3)| + 2r = 4 + 0 = 4$ (M. Stoll: Independence of rational points on twists of a given curve, Compositio Math. 142, 1201–1214 (2006)). \end{enumerate} \end{remark}

by mod 4 $x$- odd. $gcd(x,4)=1$ $x^{2}+2^2=y^{5}$ 11- 3(mod4) $y^{5}\implies 1 ,-1, mod(11)$ $11|x^2+3$ or $11|x^2+5$ contradiction