How did Letac solve $x_1^k + x_2^k + \dots +x_9^k = 0$ for $k = 1, 3, 5, 7$ in 1942?

It's quite easy to find integer solutions to,

$$x_0^k + x_1^k + \dots +x_9^k = 0$$

for $k = 1, 3, 5, 7$. One I found is, if $x^2-10y^2 = 9$, then,

$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = (-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$

However, Letac found in 1942 two special cases with $x_0 = 0$ (the case $x_0 = x_1 = 0$ is impossible), namely,

$$\{\,0, 34, 58, 82, 98\,\} = \{\,13, 16, 69, 75, 99\,\}$$

$$\{\,0, 63, 119, 161, 169\,\} = \{\,8, 50, 132, 148, 174\,\}$$

How did he do it?

The reference is A. Letac, Gazeta Matematica, 48 (1942), p. 68-69, and I'm hoping someone with JSTOR can explicitly give the answer.

Solution 1:

Of the two solutions you mention, only the second appears in Letac’s note.

He used (among other things) the following two lemmas :

Fact 1. If $a_1^k+a_2^k+a_3^k=b_1^k+b_2^k+b_3^k$ for $k=2$ and $k=4$, then $$ a_1^p+(\pm a_2)^p+(\pm a_3)^p+(2b_1)^p+(2b_2)^p+(2b_3)^p= b_1^p+(\pm b_2)^p+(\pm b_3)^p+(2a_1)^p+(2a_2)^p+(2a_3)^p \ (p=2,4,6,8) $$

Fact 2. Under the hypotheses of fact 1, for $h=\frac{a_1+a_2-a_3}{4}$ we have

$$ (2a_1-3h)^k+(2a_2-h)^k+(2a_2-3h)^k+(2a_2-h)^k+(2b_3+h)^k= (2a_3+h)^k+(2a_3+h)^k+(2b_1-h)^k+(2b_1-h)^k+(2b_2-h)^k+(3h)^k \ (k=1,3,5) $$

Solution 2:

There are some significant typos in Delanoy's answer, but I'll accept it nonetheless as it pointed me in the right direction. (Sorry for the delay.)

Let $a^k+b^k+c^k = d^k+e^k+f^k,\;\; k =2,4\tag{1}$

Theorem 1 (Birck-Sinha): "Fact 1" should be instead, given (1) then,

\begin{aligned}&(a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k + (2d)^k + (2e)^k + (2f)^k =\\ &(2a)^k + (2b)^k + (2c)^k + (-d+e+f)^k + (d-e+f)^k + (d+e-f)^k + (d+e+f)^k\end{aligned}

for $k = 2,4,6,8$.

Theorem 2 (Gloden-Sinha): "Fact 2" should be, let (1) have the constraint $a+b-c = 2(d+e-f)$, then,

\begin{aligned}&(2a-3h)^k + (2a-h)^k + (2b-3h)^k + (2b-h)^k + (2f+h)^k =\\ &(2c+h)^k + (2c+3h)^k + (2d-h)^k + (2e-h)^k + (3h)^k\end{aligned}

where $h = (d+e-f)/2$ for $k = 1,3,5,7$.

We can now reverse-engineer Letac's solution and find that he started with,

$$195^k+71^k+98^k = 21^k+190^k+127^k,\;\; k = 2,4\tag{2}$$

and using Theorem 2,

$$132^k+174^k+8^k+50^k+148^k = 119^k+161^k+0^k+169^k+63^k,\;\; k = 1,3,5,7$$

after removing a common factor. Thus, he gave (1) the second constraint $2d-h = 3d-e+f = 0$, though how he found (2) is a question in itself.

Unfortunately, it can be shown these two constraints reduce (1) to the problem of making two quartic polynomials (in one variable) into a square, hence there are only a finite number of rational solutions, and I wasn't able to find additional ones.

P.S. I'm still checking if Letac's first solution also uses Theorem 2.