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This is a duplicate question of Find integers solutions of $x^2+7=y^5$, however there was no full answer. The solutions $(\pm5, 2)$ and $(\pm 181, 8)$ have been found.

The usual strategy for such a question is to work inside the ring of integers of $\mathbb{Q}(\sqrt{-7})$, which is $\mathcal{O} = \mathbb{Z}[ \frac{1+\sqrt{-7}}{2}]$. It turns out that this is a unique factorisation domain (which one can figure out by calculating its Class group). So it is natural to factor the equation as $(x - \sqrt{-7})(x+\sqrt{-7}) = y^5$. If we assume that $x-\sqrt{-7}$ and $x+\sqrt{-7}$ are coprime, we find that $x+\sqrt{-7} = \beta^5$ for a certain $\beta = a + b\frac{1+\sqrt{-7}}{2}\in \mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$. Writing $c= 2a+b$ and expanding the fifth power, this gives the system of equations $$ c^5 -70 c^3 b^2 + 245 c^4 b = 32 x, $$ $$ 5 c^4 b -70 c^2 b^3 + 49 b^5 = 32. $$ Now with enough patience, one can show that this system has no solutions with $b \equiv c \pmod{2}$.

However this contradicts the solutions that we have found. And indeed there's no reason for $x \pm \sqrt{-7}$ to be coprime when $x$ is odd.

What is the approach to solve the remaining case of this diophantine equation?

One approach that I have tried is that the coprime condition holds inside the ring $\cal{O}[\frac{1}{2}]$. This gives the equation $x + \sqrt{-7} = (a+b\sqrt{-7})^5$ with $a,b \in \mathbb{Z}[\frac{1}{2}]$, which I am unable to solve.

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A possible factorization argument leading up till @Kevin's final form, for reference.


Since $$ (x+\sqrt{-7}) - (x-\sqrt{-7}) = 2\sqrt{-7} $$ The possible common factors are $\sqrt{-7}$ and the prime elements of norm $2$ (which works out to be $(1\pm \sqrt{-7})/2$). The former will cause $x$ to be divisible by $7$ and hence fails in the original equation, so we exclude it since it can't happen.

If $x$ is even, then $x\pm \sqrt{-7}$ has odd norm so they can't have that common factor of norm $2$. This is the part you wrote down.


However if $x=2r+1$ is odd, then since $$ \frac{x + \sqrt{-7}}{2} = r + \frac{1+\sqrt{7}}{2}\in\mathcal O, \frac{x - \sqrt{-7}}{2} = r+1 - \frac{1+\sqrt{7}}{2} \in \mathcal O, $$ the common factor between $x+\sqrt{-7}$ and $x-\sqrt{-7}$ is exactly $2$. Since we also know that $y=2s$ must be even, this means $$ \begin{align*} 2^2\left(r+\frac{1+\sqrt{-7}}{2}\right)\left(r+1-\frac{1+\sqrt{-7}}{2}\right) &= 2^5s^5\\ \left(r+\frac{1+\sqrt{-7}}{2}\right)\left(r+1-\frac{1+\sqrt{-7}}{2}\right) &= 2^3s^5 = \left(\frac{1+\sqrt{-7}}{2}\right)^3\left(\frac{1-\sqrt{-7}}{2}\right)^3s^5 \end{align*} $$ So by the coprime-ness (and absorbing any units into $\beta^5$), you now have 4 possibilities $$ r+\frac{1+\sqrt{-7}}{2} \in \left\{\beta^5,\left(\frac{1+\sqrt{-7}}{2}\right)^3\beta^5,\left(\frac{1-\sqrt{-7}}{2}\right)^3\beta^5,2^3\beta^5\right\} $$ With $\beta = a+b(1+\sqrt{-7})/2$ this works out to be 4 different (Thue) equations when comparing the real and imaginary parts: $$ \begin{align*} E1: 2r &= -1 + 2 a^5 + 5 a^4 b - 30 a^3 b^2 - 50 a^2 b^3 + 5 a b^4 + 11 b^5\\ 1 &= 5 a^4 b + 10 a^3 b^2 - 10 a^2 b^3 - 15 a b^4 - b^5\\ E2: 2r &= -1 - 5 a^5 + 5 a^4 b + 110 a^3 b^2 + 90 a^2 b^3 - 65 a b^4 - 31 b^5\\ 1 &= - a^5 - 15 a^4 b - 10 a^3 b^2 + 50 a^2 b^3 + 35 a b^4 - 3 b^5\\ E3: 2r &= -1 - 5 a^5 - 30 a^4 b + 40 a^3 b^2 + 160 a^2 b^3 + 40 a b^4 - 24 b^5\\ 1 &= a^5 - 10 a^4 b - 40 a^3 b^2 + 40 a b^4 + 8 b^5\\ E4: 2r &= -1 + 16 a^5 + 40 a^4 b - 240 a^3 b^2 - 400 a^2 b^3 + 40 a b^4 + 88 b^5\\ 1 &= 40 a^4 b + 80 a^3 b^2 - 80 a^2 b^3 - 120 a b^4 - 8 b^5 \end{align*} $$ Equation 4 clearly has no solutions modulo 2.

Equation 1 must have $b=\pm 1$, then solving for $a$ (factoring over $\mathbb Z$) gives only two integer solutions: $(a,b)= (0,-1),(1,-1)$. Then $r=-6,5$ which corresponds to $x=-11,11$, which both fails.

For equation 2, doing a substitution of $$ (a,b,r) = (-u - v, v, -w-1) $$ will reveal that it's exactly the same form as equation 3. (Upon which $(u,v,w) = (a,b,r)$ in equation 3.)


Equation 2 is given by @Kevin's solution. Alternatively Using Pari/GP to solve the Thue equation $$ 1 = - a^5 - 15 a^4 b - 10 a^3 b^2 + 50 a^2 b^3 + 35 a b^4 - 3 b^5 $$ returns $$ (a,b) = (-1, 0), (2, -1) $$ Then $r=90,2$, so $x=181,5$. Then for equation 3, using the earlier relationship of $(u,v,w) = (-a-b,b,-r-1)$ gives $w =-91,-3$. Hence $x=-181,-5$.

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    $\begingroup$ The so called Lebesgue-Nagell diophantine equation $x^2+D=y^n, n\ge3$, has been completely solved by Y. Bugeaud, M. Mignotte & S. Siksek for $1\le D\le 100$, using modular methods "à la" Wiles. See arXiv:math/0405220v1. In particular, if $D=7$, the absolute value of $x$ is $1,3,5,11$ or $181$. $\endgroup$ – nguyen quang do Jan 16 at 10:03
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Consider the general case $$x^2+7=y^m \tag{1}$$ (Integers $(x, y, m), \, m \geq 3)$ Let $$\rho = (1+\sqrt{-7})/2$$

Then as you are aware $(1, \rho)$ is a basis for the ring of integers of the field $\mathbb{Q}(-7)$. A standard factorization argument then compels us to devise a $$\frac{x-1}{2}+\rho = \rho^{m-5}(U+\rho V)^m$$

For the case where $m=5$ one needs only consider the coefficients of $(U, V)$ and form $$−U^5 − 15U^4V − 10U^3V^2 + 50U^2V^3 + 35UV^4 − 3V^5 = 1$$ This is a version of the Thue equation which I can solve using the R package NILDE.

Using this, the only solution is the only solutions to the above equation are $(U, V ) = (−1, 0),(2, −1)$, which then give solutions proper as $(5, x, y)=(5, ±5, 2),(5, ±181, 8)$

Now, in Lesage, the author shows various partial results concerning the equation $(1)$, including the following.

  • There are integer solutions to equation (1) with $m = 5, 7, 13$, this he proves by reducing to Thue equations, which he then solves by hand.
  • There are no solutions to equation (1) for $m = 11$ and for $m$ prime and $17 \leq m \leq 5000$. This he proves using classical algebraic number theory, and a computational method.
  • If $(x, y, m)$ is a solution to (1) then $m \leq 6.6 × 10^{15}$. This he proves using lower bounds for linear forms in logarithms.
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  • $\begingroup$ Could you elaborate, what is the 'standard factorisation argument' that you mention? $\endgroup$ – Ridder Jan Jan 15 at 16:49
  • $\begingroup$ Hello, are you happy with offering I gave and the answer below? $\endgroup$ – Kevin Jan 16 at 22:24

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