# Solve $a^4 + b^4 + c^4 +l^4 = a^2b^2+a^2c^2+a^2l^2+b^2c^2+b^2l^2+c^2l^2$ in integers [closed]

$$a^4 + b^4 + c^4 +l^4 = a^2b^2+a^2c^2+a^2l^2+b^2c^2+b^2l^2+c^2l^2$$ I've tried to solve this equations, have read about elliptic curves but don't know how to use it in solution. I don't know other ways to solve it.

## closed as off-topic by Carl Mummert, Namaste, Taroccoesbrocco, max_zorn, Isaac BrowneJul 27 '18 at 3:56

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• You can write a custom solution. Only do I need it? – individ Jul 24 '18 at 8:11

Geometrical solutions

Considering a point $A$ inside an equilateral $\Delta BCD$, denote $a=BC=CD=DB$, $b=AB$, $c=AC$ and $d=AD$. Applying cosine rule to those triangles, we may verify the following identity:

$$\fbox{a^4+b^4+c^4+d^4=a^2 b^2+a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2+c^2 d^2}$$

In particular when $b$, $c$ and $d$ are in AP, we have the following integral solutions:

$$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}= \begin{pmatrix} \dfrac{(2+\sqrt{3})^{2n+2}-(2-\sqrt{3})^{2n+2}}{\sqrt{3}} \\ \dfrac{(2+\sqrt{3})^{2n+2}+(2-\sqrt{3})^{2n+2}+1}{3}- \dfrac{(2+\sqrt{3})^{n+1}-(2-\sqrt{3})^{n+1}}{\sqrt{3}} \\ \dfrac{(2+\sqrt{3})^{2n+2}+(2-\sqrt{3})^{2n+2}+1}{3} \\ \dfrac{(2+\sqrt{3})^{2n+2}+(2-\sqrt{3})^{2n+2}+1}{3}+ \dfrac{(2+\sqrt{3})^{n+1}-(2-\sqrt{3})^{n+1}}{\sqrt{3}} \end{pmatrix}$$

First of all, lose the squares: $a^2=x, b^2=y, c^2=z, l^2=t$. Then the conditions are that $x,y,z,t\geq 0$ and $x^2+x^2+z^2+t^2-xy-xz-xt-yz-yt-zt=0$.

Now I am not sure what kind of solution you expect exactly, but this equations has continuum of them. Namely, if you solve for $t$, you obtain:

$$t= \frac{(x+y+z)\pm \sqrt{3}\cdot \sqrt{2xy+2yz+2zx-x^2-y^2-z^2}}{2}$$

So if the expression $2xy+2yz+2zx-x^2-y^2-z^2$ is non-negative, which again, happens in a continuum of cases, then typically there are two solutions (at least one, since choosing the plus sign always yields a solution).

So the set of points in $\mathbb{R}^4$ is the union of two 3-dimensional manifolds.

• This is not an approach to solving the equation. Because there's still a condition. $a=\sqrt{x}$ ; $b=\sqrt{y}$ ; $c=\sqrt{z}$ – individ Jul 24 '18 at 9:52
• I can see now that you meant INTEGER solutions, but this can only be implied from the tags "number theory" and "integers". Please include the condition in the post! – A. Pongrácz Jul 24 '18 at 11:37

$$a^4+b^4+c^4+q^4=a^2b^2+a^2c^2+a^2q^2+b^2c^2+b^2q^2+c^2q^2$$

You can record such a parameterization of solutions.

$$a=p^8-16p^7s+104p^6s^2-356p^5s^3+709p^4s^4-864p^3s^5+666p^2s^6-324ps^7+81s^8$$

$$b=p^8-12p^7s+74p^6s^2-288p^5s^3+709p^4s^4-1068p^3s^5+936p^2s^6-432ps^7+81s^8$$

$$c=p^8-8p^7s+20p^6s^2-4p^5s^3-41p^4s^4-12p^3s^5+180p^2s^6-216ps^7+81s^8$$

$$q=4ps(p^2-4ps+3s^2)(3s-2p)(p-2s)(p^2-3ps+3s^2)$$