How to prove that the Diophantine equation $$(x+y)(x+y+2)=10xy\quad (1)$$ has no positive integer solutions

First attempt as below is wrong :is to rewrite this equation as $$y^2 + y(2-8x) + x^2 + 2x = 0$$. The discriminant is $\Delta = 4(15x^2 - 10x + 1)$. Therefore, $y = 4x - 1 \pm \sqrt{15x^2 - 10x + 1}$. If a solution of the equation (1) exists for some integers $x, y > 0$, then $15x^2 - 10x + 1$ must be a perfect square. However, its discriminant is not zero, hence the equation (1) has no positive integer solutions for $x$ and $y$.

Second adempt : $15x^2 - 10x + 1$ must be a perfect square, Wolfram gives the solution of the equation $15x^2 - 10x + 1=m^2$. We can remark that all $x$ solutions are <0, which shows that no positive solution can exist. But can we retrieve Wolfram's results? I don't know.

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    $\begingroup$ It could still be possible for $P(n)$ to be the square of some integer despite there not being a polynomial $Q\in\Bbb C[x]$ such that $P=Q^2$. For instance, $P(x)=x^2+5$ does that for $x=2\lor x=-2$. $\endgroup$ Apr 10 at 3:01
  • $\begingroup$ @SassatelliGiulio You are right $\endgroup$
    – Noname
    Apr 10 at 3:09
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    $\begingroup$ Thank you and maybe someone who knows these Pell Fermat equations well will tell us one more $\endgroup$
    – Noname
    Apr 10 at 20:27
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    $\begingroup$ @JohnOmielan, you can find the origin of this question in this thread: [mathoverflow.net/questions/468844/…. However, it's closed $\endgroup$
    – Noname
    Apr 11 at 17:46
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    $\begingroup$ new link math.stackexchange.com/questions/4897488/… $\endgroup$
    – Noname
    Apr 11 at 23:57

2 Answers 2


The Diophantine equation to check on is

$$(x + y)(x + y + 2) = 10xy \tag{1}\label{eq1A}$$

Note that, by inspection, I got $(x, y) \in \{(0, 0), (0, -2), (-2, 0)\}$ being integer solutions, but not positive ones. Since there are integer solutions, doing things like factoring and checking for a perfect square discriminant will succeed (e.g., with $15x^2 - 10x + 1$, it's $1$ for $x = 0$ and $81 = 9^2$ for $x = -2$, with your Wolfram Alpha results show other, only negative, results)). One way to specifically show there are no positive integer solutions is to assume they exist but then use positive divisibility constraints (which assume all of the factors are non-negative), such as with my \eqref{eq4A}, \eqref{eq5A} and \eqref{eq6A} below, to show there are no valid results.

First, since $x + y$ and $x + y + 2$ have the same parity and $10xy$ is even, then both of the LHS factors of \eqref{eq1A} must be even. Thus, $x$ and $y$ have the same parity. They can't both be odd since the LHS has at least $2$ factors of $2$ but the RHS would have only $1$ factor of $2$. This means $x$ and $y$ must both be even. As such, there's a positive integer $d$ where

$$\gcd(x, y) = 2d \;\;\to\;\; x = 2dx_1, \; y = 2dy_1, \; \gcd(x_1, y_1) = 1 \tag{2}\label{eq2A}$$

Substituting this into \eqref{eq1A} and dividing both sides by $4d$ gives

$$(x_1 + y_1)(dx_1 + dy_1 + 1) = 10dx_{1}y_{1} \tag{3}\label{eq3A}$$

Since $d \mid (x_1 + y_1)(dx_1 + dy_1 + 1)$, but $\gcd(d, dx_1 + dy_1 + 1) = 1$, we have

$$d \mid x_1 + y_1 \tag{4}\label{eq4A}$$

Also, $x_1 + y_1 \mid 10d(x_{1}y_{1})$, but $\gcd(x_1, y_1) = 1$ means that $\gcd(x_1 + y_1, x_{1}y_{1}) = 1$, so

$$x_1 + y_1 \mid 10d \tag{5}\label{eq5A}$$

Note that \eqref{eq4A} and \eqref{eq5A}, along with \eqref{eq3A}, means there are positive integers $a$ and $b$ with

$$x_1 + y_1 = ad, \;\; d(x_1 + y_1) + 1 = bx_{1}y_{1}, \;\; ab = 10 \tag{6}\label{eq6A}$$

Substituting the LHS of \eqref{eq6A} into the middle part gives

$$\begin{equation}\begin{aligned} d(ad) + 1 & = b(ad - y_1)y_1 \\ ad^2 + 1 & = (ab)dy_1 - by_1^2 \\ by_1^2 - 10dy_1 + (ad^2 + 1) & = 0 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Treating this as a quadratic in $y_1$, the discriminant, i.e.,

$$100d^2 - 4b(ad^2 + 1) = 100d^2 - 4(10)d^2 - 4b = 60d^2 - 4b = 4(15d^2 - b) \tag{8}\label{eq8A}$$

must be a perfect square. Thus,

$$15d^2 - b \tag{9}\label{eq9A}$$

must also be a perfect square. From the RHS of \eqref{eq6A}, we get these cases to check for $b$:

  1. Since $15d^2 - 1 \equiv 2\pmod{3}$, it's not a perfect square. In addition, if $d$ is odd, then $15d^2 - 1 \equiv 7(1) - 1 \equiv 6 \pmod{8}$, while if $d$ is even, then $15d^2 - 1 \equiv 3 \pmod{4}$, with neither being possible.

  2. With $15d^2 - 2 \equiv 3 \pmod{5}$, it can't be a perfect square. Another method is to note that $d$ can't be even since it would have only $1$ factor of $2$, so $d$ is odd. However, then $15d^2 - 2 \equiv 7(1) - 2 \equiv 5 \pmod{8}$.

  3. Checking $15d^2 - 5$, we get $d$ can't be even because then $15d^2 - 5 \equiv 3 \pmod{4}$. However, $d$ being odd means $15d^2 - 5 \equiv 7(1) - 5 \equiv 2 \pmod{8}$, which is also never true for perfect squares. Alternatively, $15d^2 - 5 = 5(3d^2 - 1)$, so $5 \mid 3d^2 - 1$. However, $d^2 \equiv 0, 1, 4 \pmod{5}$, with $5 \nmid 3d^2 - 1$ for each congruence.

  4. Finally, $15d^2 - 10 \equiv 2 \pmod{3}$, so this can't be a perfect square either. Another way to show this is that $d$ can't be even because $15d^2 - 10$ would have only one factor of $2$, but $d$ being odd means that $15d^2 - 10 \equiv 7(1) - 2 \equiv 5 \pmod{8}$.

Since these all result in a non-perfect square discriminant, there are no positive integer solutions for $d$ and $y_1$ in \eqref{eq7A} and, thus, also for $x$ and $y$ in \eqref{eq1A}.

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    $\begingroup$ Thank you for this very detailed proof. I will study this solution. I'll come back later to accept it or ask for clarification if I get stuck on any obscure points. $\endgroup$
    – Noname
    Apr 10 at 8:33
  • $\begingroup$ @Noname You're welcome. Note I've updated my answer, including adding a few more details, that you may find helpful. $\endgroup$ Apr 10 at 17:53
  • $\begingroup$ @Noname Easier way: use descent on integer points on the conic ("Vieta jumping"), cf. my comment on the question. $\endgroup$ Apr 13 at 12:24
  • $\begingroup$ Hello, @Bill Dubuque 1 , could you please share your solution to let readers of the thread learn from your approach? Additionally, if someone asks me (now that I understand how to solve a generalized Pell-Fermat equation), I can explain the results from Wolfram. $\endgroup$
    – Noname
    Apr 13 at 13:41
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    $\begingroup$ I'm also saddened to hear about your health issue. I wish you strength and courage to fight your illness. $\endgroup$
    – Noname
    Apr 14 at 17:29

The resulting Diophantine equation is

$$y^2 + y(2 - 8x) + x^2 + 2x = 0 \tag{1}\label{eq1B}$$

As indicated in Bill Dubuque's comment, we can use Standard Vieta jumping to prove there are no positive integer solutions. Similar to what's shown there with the example Problem #$6$ at IMO $1988$, let $(X,Y)$ be a solution in positive integers to \eqref{eq1B}, i.e.,

$$Y^2 + Y(2 - 8X) + X^2 + 2X = 0 \tag{2}\label{eq2B}$$

such that $X + Y$ is minimized and, due to symmetry, WLOG let $Y \ge X$. Actually, since $Y = X$ in \eqref{eq2B} leads to $X = 0$ or $X = \frac{2}{3}$, we can instead have $Y \gt X$. Even more strongly, as I explained in my other answer, due to parity and # of factors of $2$, both $X$ and $Y$ are even, so we then get

$$Y \ge X + 2 \;\to\; \frac{1}{Y} \le \frac{1}{X + 2} \tag{3}\label{eq3B}$$

Fixing $X$, replace $Y$ with the variable $z$ in \eqref{eq2B} to get

$$z^2 + z(2 - 8X) + X^2 + 2X = 0 \tag{4}\label{eq4B}$$

We know one root of this equation is $z_1 = Y$. Using the standard properties of quadratic equations, the other root satisfies

$$z_2 = -(2 - 8X) - Y, \;\; z_2 = \frac{X^2 + 2X}{Y} \tag{5}\label{eq5B}$$

The first expression shows that $z_2$ is an integer, while the second indicates it's positive. Using the second expression, and \eqref{eq3B}, we have

$$z_2 = \frac{X^2 + 2X}{Y} = X(X + 2)\left(\frac{1}{Y}\right) \le X(X + 2)\left(\frac{1}{X + 2}\right) = X \tag{6}\label{eq6B}$$

However, then

$$z_2 \le X \;\;\to\;\; z_2 \lt Y \;\;\to\;\; X + z_2 \lt X + Y \tag{7}\label{eq7B}$$

contradicting the minimality of $X + Y$ (since $(x, y) = (X, z_2)$ is another solution in positive integers to \eqref{eq1B}). This contradiction proves there are no positive integer solutions to \eqref{eq1B}.

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    $\begingroup$ +1 This is essentially the algebra behind the method I hinted in comments. I encourage readers to geometrically visualize the descent by graphing it - draw a zigzag line between each pair of points in the descent, generated by the reflections $\,(x,y) \to (y,x)\,$ and the "conjugate root" reflection $\,(x,y)\to (x,\bar y),\,$ where $\,\bar y = 8x\!-\!2-y = (x^2\!+2x)/y,\,$ by Vieta. For general theory see the literature I cite in this answer. $\endgroup$ Apr 14 at 5:45
  • $\begingroup$ To elaborate, while the descent on integer points on the positive branch is only hypothetical in the proof (since there are no positive integer points), it actually does exist on the lower negative branch, e.g. $(-18,-144)\to(-18,-2)\to(-2,-18)\to(-2,0)\to(0,-2)\to(0,0)$. It is instructive to analyze how this differs from the positive branch, and how it can be remedied by scaling $x,y$ by $3$ so our positive fixed point $(2/3,2/3)$ becomes integral $= (2,2).\,$ Now the positive branch of our scaled curve $\,x^2-8xy+y^2+6x+6y=0$ has integer points e.g. $(8,56)\to(8,2)\to(2,8)\to(2,2)\ \ $ $\endgroup$ Apr 14 at 6:15
  • $\begingroup$ If someone is handy with graphic software it would be very helpful to readers to add graphs illustrating these points. $\endgroup$ Apr 14 at 6:19

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