Exercise 5.4.8 (Diophantine equation with $18$ variables)

Question

Let $f({\bf x}) = f(x_1,x_2,x_3) = x_1^4+x_2^4+x_3^4 - x_1^2x_2^2 - x_2^2x_3^2 - x_3^2x_1^2 - x_1x_2x_3(x_1+x_2+x_3)$. Show that $f({\bf x}) \equiv 1 \pmod 4$ unless all three variables are even. Deduce that if $f({\bf x})+ f({\bf y})+ f({\bf z}) = 4 (f({\bf u}) + f({\bf v})+ f({\bf w}))$ for integral values of the variables, then all $18$ variables are $0$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
If all three variables are even, then $ f(x_1,x_2,x_3) \equiv 0 \pmod 4$. It remains $56 = 64 - 8$ verifications with Sagemath
\begin{verbatim}
A = set()
for x1 in Integers(4):
    for x2 in Integers(4):
        for x3 in Integers(4):
            r = x1^4 + x2^4 + x3^4 - x1^2 * x2^2 - x2^2 * x3^2 -
                   x3^2 * x1^2 - x1 * x2 * x3 * (x1 + x2 + x3)
            if x1.lift() % 2 != 0 or x2.lift() % 2 != 0 or x3.lift() % 2 != 0:
                A.add(r)
A
    {1}
\end{verbatim}
This shows that $f(x_1,x_2,x_3) \equiv 1 \pmod 4$ unless all three variables are even.

\bigskip
Suppose for the sake of contradiction that 
\begin{align}
f(x_1,x_2,x_3) + f(y_1,y_2,y_3) + f(z_1, z_2,z_3) = 4( f(u_1,u_2,u_3) + f(v_1,v_2,v_3)+ f(w_1,w_2,w_3)),
\end{align}
where $(x_1,x_2,x_3, y_1,y_2,y_3, z_1, z_2,z_3,u_1,u_2,u_3, v_1,v_2,v_3, w_1,w_2,w_3) 
e (0,0,\ldots,0)$.

If we divide by $d^4$, where $d$ is the gcd of these $18$ values, we obtain a similar equation, where the new $18$ values are relatively prime. So we can assume that in (1),
$$x_1\wedge x_2\wedge x_3,\wedge y_1\wedge y_2  \wedge y_3 \wedge z_1\wedge  z_2 \wedge z_3 \wedge u_1 \wedge u_2 \wedge u_3 \wedge v_1 \wedge v_2 \wedge v_3 \wedge w_1 \wedge w_2 \wedge w_3 = 1.$$
(We use the same Fermat's method of descent as in the previous problems.)

Reducing modulo 4, we obtain
$$f(x_1,x_2,x_3) + f(y_1,y_2,y_3) + f(z_1, z_2,z_3) \equiv 0 \pmod 4.$$
Since $f(x_1,x_2,x_3) \equiv 0 	ext{ or } 1 \pmod 4$, this congruence implies
$$f(x_1,x_2,x_3) \equiv  f(y_1,y_2,y_3)\equiv  f(z_1, z_2,z_3) \equiv 0 \pmod 4.$$
By the first part,
$$x_1 \equiv x_2 \equiv x_3 \equiv y_1 \equiv y_2 \equiv y_3 \equiv z_1 \equiv z_2 \equiv z_3 \equiv 0 \pmod 2.$$
Then $2^4 \mid ff(x_1,x_2,x_3) + f(y_1,y_2,y_3) + f(z_1, z_2,z_3)$, thus by (1), $2^4 \mid 4( f(u_1,u_2,u_3) + f(v_1,v_2,v_3)+ f(w_1,w_2,w_3))$. This shows that
$$f(u_1,u_2,u_3) + f(v_1,v_2,v_3)+ f(w_1,w_2,w_3) \equiv 0 \pmod 4.$$
By the same reasoning,
$$ u_1 \equiv u_2\equiv u_3 \equiv v_1 \equiv v_2 \equiv v_3 \equiv w_1 \equiv w_2 \equiv w_3 \equiv 0 \pmod 2.$$
Therefore
$$2 \mid x_1\wedge x_2\wedge x_3,\wedge y_1\wedge y_2  \wedge y_3 \wedge z_1\wedge  z_2 \wedge z_3 \wedge u_1 \wedge u_2 \wedge u_3 \wedge v_1 \wedge v_2 \wedge v_3 \wedge w_1 \wedge w_2 \wedge w_3 = 1.$$
This contradiction $2 \mid 1$ shows that the equation $f(x_1,x_2,x_3) + f(y_1,y_2,y_3) + f(z_1, z_2,z_3) = 4( f(u_1,u_2,u_3) + f(v_1,v_2,v_3)+ f(w_1,w_2,w_3))$ has no nontrivial integral solution.

\end{proof}

Exercise 5.4.8 (Diophantine equation with $18$ variables)

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Exercise 5.4.8 (Diophantine equation with $18$ variables)

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