Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers.

Question

Several formulae that solve the diophantine equation $$w^4 + x^4 = y^4+z^4 \tag{1}$$ are presented in this collection. The simplest one bases on $$f_1 = a^7 + a^5 - 2 a^3 + 3 a^2 + a \tag{2}$$ and sets$\def\hf{{}^h\!f}$ $$ (w, x, y, z) = (\hf_1,\ \hf_1(b,-a),\ \hf_1(a,-b),\ \hf_1(b,a)) $$ where $\hf(a,b) = f(a/b)\cdot b^{\deg f}$ denotes the homogenized version of $f$.

Question: How does one find such formulae?

The collection refers to Hardy & Wright, and MathWorld mentions that parametric solutions to (1) were already known to Euler in 1802, but without mentioning which one.

Note: $f_1$ appears to be — in some sense — the simplest of such functions: It produces (158, 59, 134, 133) from (1,2), the smallest solution of (1). All solutions that I checked using Sage were non-trivial except for the case $a=b=1$, and with $\gcd(a,b)=1$ all solutions satisfied $\gcd(w,x,y,z)\in\{1,2\}$. The last two properties also showed up for the next 2 more complicted functions from the collection that I tried like $$f_2 = -a^{13} + a^{12} + a^{11} + 5 a^{10} + 6 a^{9} - 12 a^{8} - 4 a^{7} + 7 a^{6} - 3 a^{5} - 3 a^{4} + 4 a^{3} + 2 a^{2} - a + 1 $$

Answer 1

$$w^4 + x^4 = y^4 + z^4 \tag{1}$$

Let $w=mt+a, x=t-b, y=t+a, z=mt+b,$ then we get

$(-4a-4b-4bm^3+4am^3)t^3+(-6b^2m^2-6a^2+6b^2+6a^2m^2)t^2+(-4b^3m+4a^3m-4a^3-4b^3)t=0$

Hence let $$m = \frac{a^3+b^3}{-b^3+a^3}$$ then we get $$t = \frac{3(-b^3+a^3)b^2a^2}{a^6-2b^2a^4-2b^4a^2+b^6}$$

Thus we get a parametric solution as follows.

$w = a(a^6+b^2a^4-2b^4a^2+3b^5a+b^6)$
$x = b(a^6-3ba^5-2b^2a^4+b^4a^2+b^6)$
$y = a(a^6+a^4b^2-2a^2b^4-3ab^5+b^6)$
$z = b(a^6+3ba^5-2b^2a^4+b^4a^2+b^6)$

Let $b=1$, we get

$w = a^7+a^5-2a^3+3a^2+a$
$x = a^6-3a^5-2a^4+a^2+1$
$y = a^7+a^5-2a^3-3a^2+a$
$z = a^6+3a^5-2a^4+a^2+1$

This solution is not a complete solution.

Several solutions are found on A^4+B^4=C^4+D^4

Answer 2

The point here is to note that the equation $$x^4 + w^4 - y^4 - z^4 = 0$$ cuts out a surface $X \subset \mathbb{P}^3$. Indeed, all (primitive, i.e., $\operatorname{gcd}(x,y,z,w) = 1$) integral solutions are in bijective correspondence with rational points on the surface $X$.

Now, one might hope for a solution with $2$ parameters (i.e., a birational map $\mathbb{P}^2 \to X$ defined over $\mathbb{Q}$) however, this is not the case. One may see this since $X$ is a nonsingular quartic surface, it is a $K3$ surface - so is not birational to $\mathbb{P}^2$.

Instead, the best one can hope for in terms of parametric solutions is to find a curve $C \subset X$ lying on $X$ which has genus $0$. Since in this case we may parametrise $C$ and find a $1$-parameter family of solutions to our original equation.

There are some obvious curves that you already know about. For example the curve given by $x=y$ and $w=z$ is pretty stupid, but you do get the parametrisation $(w,x,y,z) = (1, a, a, 1)$, which is a perfectly legitimate option, but I presume that you are calling these 'trivial' solutions.

I'm not going to regurgitate @Tomita's answer, which shows how to construct the particular genus $0$ curve you're interested in which lies on $X$.

In general it is an interesting question to find genus $0$ curves on a $K3$ surface. If $X$ is elliptic $K3$ then one can find a fibration and use the theory of elliptic curves over function fields, otherwise I am not sure of any general theory which allows one to do this.

Answer 3

From L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXII, page $644$

$$ A^4+B^4=C^4+D^4. $$ L. Euler$^{165}$ took $\,A=p+q,\;\;D=p-q,\;\;C=r+s,\;\;B=r-s\,$ and derived $$(1)\qquad\qquad\qquad pq(p^2+q^2)=rs(r^2+s^2).$$ Set $\,p=ax,\,q=by,\,r=kx,\,s=y.\,$ Then $$ y^2/x^2=(k^3-a^3b)/(ab^3-k). $$ If $\,k=ab,\,x=1,\,$ then $\,y=\pm a,\,C=\pm A,\,B=\mp D.\,$ Set therefore $\,k=ab(1+z).\,$ Then $\, y^2/x^2=a^2Q/(b^2-1-z)^2,\,$ where $$ Q\!=\!(b^2\!-\!1)^2\!+\!(b^2\!-\!1)(3b^2-1)z \!+\! 3b^2(b^2\!-\!2)z^2\!+\!b^2(b^2\!-\!4)z^3\!-\!b^2z^4. $$ Let $\,Q\,$ be the square of $\,b^2-1+fz+gz^2\,$ and choose $\,f,g\,$ to make the terms in $\,z,z^2\,$ agree. Thus, $$ f=\frac{3b^2-1}2,\quad g=\frac{3b^4-18b^2-1}{8(b^2-1)},\quad z=\frac{b^2(b^2-4)-2fg}{b^2+g^2}. $$ Then $\,x:y=b^2-1-z:a(b^2-1+fz+gz^2).\,$ As examples, Euler took $\,b=2,\,b=3\,$ and found the solution $$ A=2219449,\quad B=-555617,\quad C=1584749,\quad D=2061283,$$ and an erroneous$^{166}$ one replaced in his next paper by $$ A=12231,\quad B=2903,\quad C=10381,\quad D=10203.$$

I have verified the steps for Euler as quoted in Dickson. For $\,b=1\,$ the result is $\,A=1,\,B=-1,\,C=1,\,D=-1\,$ and for $\,b=2,\,b=3\,$ the results are as given.

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The general result given by Euler depends on one parameter $\,b.\,$ Define the two variable polynomial $$ P(u,v) := u^{13}+27 u^{12} v-214 u^{11} v^2-186 u^{10} v^3- 2481 u^9 v^4+861 u^8 v^5-2804 u^7 v^6\\-972 u^6 v^7-2481 u^5 v^8- 27 u^4 v^9-214 u^3 v^{10}+294 u^2 v^{11}+u v^{12}+3 v^{13}. $$ The solution given by Euler is essentially $$ A=P(-b,1),\quad B=P(1,b),\quad C=-P(1,-b),\quad D=P(b,1).$$ Note that $\,P(u,v) = {}^h\!f_2(u+v,v-u)\,$ is the connection with your solution $\,f_2.$

The polynomial corresponding to your $\,f_1\,$ is $$ P(u,v) := u^6 v+3 u^5 v^2-2 u^4 v^3+u^2 v^5+v^7 $$ due to Gerardin which I got from A Collection of Algebraic Identities by Tito Piezas III.

Dickson mentions Gerardin briefly on page 647

A. Gerardin$^{181a}$ noted that $(1)$ has the solution $$ p=a^7+a^5-2a^3+a,\quad q=3a^2,\quad r=a^6-2a^4+a^2+1,\quad s=3a^5,$$ which is simpler than Euler's solution $(5)$.

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Your question

How does one find such formulae?

is a difficult one. Another answer mentions $K3$ surfaces of which I know almost nothing. Particular cases of interest include $\,A^4+B^4+C^4+D^4=0\,$ which is the Fermat quartic. The article Euler's sum of powers conjecture mentions the case $\,A^4+B^4+C^4=D^4\,$ which was solved by Noam Elkies using elliptic curves.