Problem A. 858. (September 2023)

A. 858. Prove that the only integer solution of the following system of equations is \(\displaystyle u=v=x=y=z=0\): \(\displaystyle uv=x^2-5y^2\), \(\displaystyle (u+v)(u+2v)=x^2-5z^2\).

Proposed by Barnabás Szabó, Budapest

(7 pont)

Deadline expired on October 10, 2023.

Suppose for the sake of contradiction that there exists a solution where not all variables are equal to 0. Let's pick one of these solutions where the sum of the absolute values is the smallest possible.

First we show that none of \(\displaystyle u\), \(\displaystyle v\), \(\displaystyle u+v\) and \(\displaystyle u+2v\) can be divisible by 5.

If at least one of \(\displaystyle u\), \(\displaystyle v\), \(\displaystyle u+v\) and \(\displaystyle u+2v\) is divisible by 5, then \(\displaystyle x\) is divisible by 5. Thus the right hand side of both equations is divisible by 5, and then at least two of \(\displaystyle u\), \(\displaystyle v\), \(\displaystyle u+v\) and \(\displaystyle u+2v\) is divisible by 5. A quick check shows that in this case both \(\displaystyle u\) and \(\displaystyle v\) must be divisible by 5, and thus the left hand sides of the equations are divisible by 25. This implies \(\displaystyle y\) and \(\displaystyle z\) must also be divisible by 5. However, this makes it possible to divide all variables by 5, and then we get a solution with a smaller sum of the absolute values, which contradicts our assumption. Thus none of \(\displaystyle u\), \(\displaystyle v\), \(\displaystyle u+v\) and \(\displaystyle u+2v\) is divisible by 5.

The next claim is the key to the solution: if \(\displaystyle p\) is an odd prime that divides \(\displaystyle x^2-5y^2\) and \(\displaystyle p\) does not divide \(\displaystyle x\) and \(\displaystyle y\), then \(\displaystyle p\) is 1 or -1 mod 5. The reason is that this is equivalent to 5 being a quadratic remainder mod \(\displaystyle p\) (being the square of \(\displaystyle x/y\)), and by applying quadratic reciprocity \(\displaystyle p\) is a quadratic remainder mod 5 since 5 is a prime that is 1 mod 4. Thus \(\displaystyle p\) must be 1 or -1 mod 5.

This implies the following: if \(\displaystyle p\) is odd and it is 2 or 3 mod 5, then the exponent of \(\displaystyle p\) in \(\displaystyle x^2-5y^2\) is even. And this means that all powers of primes different from 5 in \(\displaystyle x^2-5y^2\) will be 1 or -1 mod 5. Notice that the same is also true for \(\displaystyle 2\): if \(\displaystyle x\) and \(\displaystyle y\) are both odd, then \(\displaystyle x^2-5y^2\) is 4 mod 8, so the exponent of 2 is 2, and if both are even, we can divide them by 2, and \(\displaystyle x^2-5y^2\) will be divided by 4.

Now if both \(\displaystyle u\) and \(\displaystyle v\) are divisible by \(\displaystyle p\neq 5\), then all variables could be divided by \(\displaystyle p\), which is a contradiction. If only one of them is divisible by \(\displaystyle p\), then according to the above the power of \(\displaystyle p\) in the prime factorization of it will be 1 or -1 mod 5.

If both \(\displaystyle u+v\) and \(\displaystyle u+2v\) are divisible \(\displaystyle p\neq 5\), then both \(\displaystyle u\) and \(\displaystyle v\) are divisible by \(\displaystyle p\), and we have seen that this is not possible. Thus the same is true for \(\displaystyle u+v\) and \(\displaystyle u+2v\) as before: each prime power in their prime factorization has to be 1 or -1 mod 5.

Since none of \(\displaystyle u\), \(\displaystyle v\), \(\displaystyle u+v\) and \(\displaystyle u+2v\) can be divisible by 5, the above means that all of them has to be 1 or -1 mod \(\displaystyle 5\). However, it is really easy to check that this cannot happen even for \(\displaystyle u\), \(\displaystyle v\) and \(\displaystyle u+v\), and thus we are done.

Statistics:

19 students sent a solution.
7 points:	Bodor Mátyás, Diaconescu Tashi, Duchon Márton, Selim Cadîr, Simon László Bence, Varga Boldizsár, Wiener Anna.
6 points:	Lincoln Liu.
2 points:	1 student.
1 point:	2 students.
0 point:	8 students.

Problems in Mathematics of KöMaL, September 2023