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Find all triples of integers $(x,y,d)$ with $d\ge 3$ such that $x^2+4=y^d$.

I did some advance in the problem with Gaussian integers but still can't finish it. The problem is similar to Catalan's conjecture.

NOTE: You can suppose that $d$ is a prime.

Source: My head

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Trivial observation: if $d$ is even there is no nontrivial solution. –  N. S. Mar 11 '12 at 17:37
I'm wondering, why is the tag (complex-numbers) used here? The problem is asking for integer solutions to an integer polynomial. –  user2468 Mar 11 '12 at 17:47
@J.D. Probably b/c OP approached it using $\mathbb Z[i]$. –  Alex Becker Mar 11 '12 at 17:48
The case $d=3$ is solved as Example 4.1.4 in the book An Introduction to Diophantine Equations: A Problem-Based Approach by Titu Andreescu,Dorin Andrica,Ion Cucurezeanu, p. 157. –  Martin Sleziak Mar 26 '12 at 14:51

6 Answers 6

What follows only takes care of $d=3$. We use the Gaussian integer approach that you mentioned. The method can be used with other $d$, but some details of the calculation depend on $d$. Thus one can only deal with one $d$ at a time.

We first examine the case $x$ odd. Factor $x^2+4$ in the Gaussian integers as $(x+2i)(x-2i)$. Since $x$ is odd, $x+2i$ and $x-2i$ are relatively prime. If their product is a perfect cube, each must be a (Gaussian) perfect cube. Here we are using the Unique Factorization Theorem for $\mathbb{Z}[i]$. Some care needs to be taken, because of units that may appear in front of the prime factorization. But since all four units are cubes, they give no difficulty.

Let $x+2i=(a+bi)^3$. Expand, and compare real and imaginary parts. We get $a^3-3ab^2=x$ and $3a^2b-b^3=2$. Since $3a^2b-b^3=b(3a^2-b^2)$, there are only the possibilities $b=\pm 1$, $b=\pm 2$. The case $b=1$ gives $a=\pm 1$. The case $b=-1$ gives no solution. The case $b=2$ also gives no solution, while $b=-2$ gives $a=\pm 1$. The case $b=1$ gives the solution $x=2$ which we were not looking for, and $b=-2$, $a=-1$ gives the solution $x=11$.

Next we deal somewhat more briefly with the case $x$ even. Let $x=2s$ and $z=2t$. Our equation becomes $(s+i)(s-i)=2t^3$. Note that $s$ and $t$ must be odd. By considering factorizations again, we get that $s+i$ must have shape $(1\pm i)(a+bi)^3$, and an analysis similar to the one above works. We get (for $(1+i)(a+bi)^3$) that $(a-b)(a^2+4ab+b^2)=1$, which forces $b=0$, $a=1$ or $b=-1$, $a=0$. Thus the only positive solution of $s^2+1=2t^3$ has $s=1$, and therefore the only positive even $x$ such that $x^2+4$ is a perfect cube is given by $x=2$.

Remark: Let $d>3$ be odd. We could, for any specific $d$, use exactly the same strategy. Think for example of $d=5$, and the case $x$ odd. Then from the fact that $x+2i$ must be a perfect (Gaussian) fifth power, we get $x=a^5-10a^3b^2+5ab^4$ and $2=5a^4b-10a^2b^3-b^5$. The second equation forces $b=\pm 1$ or $b=\pm 2$. For these four possibilities we can quickly find any integer solution $a$ by using the "Factor Theorem." If we get an integer solution $a$, that gives a solution of the original equation, and if we don't, we know the original equation has no solution.

For any fixed odd $d$, the kind of calculation described above can be carried out mechanically and quickly. And it is clear from the analysis that for any fixed $d$, there can be at most finitely many solutions. But unfortunately the strategy can only take care of $d$'s one at a time.

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I believe the only solutions are $(\pm 2)^2 + 4 = 2^3$ and $(\pm 11)^2 + 4 = 5^3$. See https://oeis.org/A076427 and references given there.

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Not the I'm suggesting anything, but I did an experiment.

For $3 < d \le 30, 1 \le y \le 1000$, this expression $\sqrt[2]{y^d - 4}$ is not an integer. Only when $d = 3$, the solutions are $x = \pm 2$, and $\pm 11$. But again, ranges are small and this proves nothing.

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Assume that there exist $x, y, d \in \mathbb{Z}$ with $d \geq 3$ and $x$ odd, satisfying $x^2 + 4 = y^d$. Then the ideal $\mathfrak{a} = (y, x + 2 i)$ of $\mathbb{Z}[i]$ satisfies $\mathfrak{a}^d$ is principal. However, since $\mathbb{Z}[i]$ has unique factorization, the ideal $(y, x + 2 i)$ is principal irrespective of $d$. Thus we are looking for $a, b \in \mathbb{Z}$ such that $(y, x + 2 i) = (a + b i)$. For example, $(5, 11 +2 i) = (2- i)$. Maybe you can at least impose some restrictions on $x, y$.

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Here's a proof sketch. It's missing two major steps Propositions 1 and 2, but I'll try to fill them in and come back with an update.

First consider $x$ and $y$ even. Then we can only have $x\equiv 4\pmod{16}$, $y\equiv 8\pmod{16}$ and hence that $d=3$, for which @André Nicolas showed that $x=y=2$ is the only solution.

Now consider $y$ odd. $d$ must be odd, so we can rewrite the equation as $$ yz^2-x^2=4 $$ When $y=a^2+4$ then this has solutions $$ \begin{array}{l} z_0=z_1=1,z_{k+2}=(y-2)z_{k+1}-z_k \\ x_0=-a,x_1=a,x_{k+2}=(y-2)x_{k+1}-x_k \end{array} $$

Proposition 1: These enumerate all solutions, and there are no solutions in integers when $\sqrt{y-4}$ is not an integer.

Now to find a solution to the original problem we would need $z_i$ to be a power of $y$ for some $i$. We can look at $z_k$ modulo $y$ and modulo $y^2$: $$ \begin{align} z_k & \equiv (-1)^{k+1}(2k-1) \pmod{y}\\ z_k & \equiv (-1)^{k}(2k-1)\left(\frac{k(k-1)}{6}y-1\right) \pmod{y^2}\\ \end{align} $$ (These can be shown by induction.)

Thus $y\mid z_k$ iff $y\mid 2k-1$ and the first $z_k$ divisible by $y$ is when $k=(y+1)/2$.

Proposition 2: If $y\mid z_k$ then $z_{(y+1)/2} \mid z_k$.

Taking $z_{(y+1)/2}$ mod $y^2$:

$z_{(y+1)/2} \equiv -y\left(\frac{y^2-1}{24}y-1\right) \equiv -yw \pmod{y^2} $

where $\gcd(y,w)=1$. Hence $z_{(y+1)/2}$ is never divisible by $y^2$ and is either equal to $y$ or has a factor greater than 1 that does not divide $y$.

Since $z_k$ is increasing in $k$ and $z_3 = y^2-5y+5$ we find that $z_3=5$ when $y=5$ and $y<z_3<z_{(y+1)/2}$ when $y>5$. Thus $z_{(y+1)/2}$ is not a power of $y$ for $y>5$, and assuming Prop 2 nor can any other $z_k$ be, so a solution can only exist if $y=5$.

If $y=5$, $z_3=5,d=3,x=11$ is one solution.

The case $y=5$ is special in that $z_k=F_{2k-1}$ where $F_n$ is the $n$th Fibonacci number. We can use the divisibility properties of $F_n$ to show that if $25\mid z_k$ then $3001 \mid F_{25} \mid z_k$ so $z_k$ is not a power of 5. (This also shows Prop 2 is true when $y=5$ and hints how it might be proved in general.)

Thus the only solutions are $(x,y,d)=(\pm 2,2,3)$ or $(x,y,d)=(\pm 11,5,3)$.

Edit: Proposition 1 is incorrect, there may be solutions when $\sqrt{y-4}$ is not an integer, e.g. $39^2 = 61\cdot5^2-4$. This looks to be fatal to the argument.

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See a similar question that I asked recently: Nontrivial Rational solutions to $y^2=4 x^n + 1$

This question might also be related to Fermat's Last Theorem.

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