# A Turkish number theory problem for RMO

(Turkey 1997)

Prove that for each prime $p \ge 7$, there exists a positive integer n and integers $x_{1}, \ldots, x_{n}$, $y_{1},\ldots, y_{n}$ not divisible by p such that $x_{1}^{2}+y_{1}^{2} \equiv x_{2}^{2} \pmod p$ $x_{2}^{2}+y_{2}^{2} \equiv x_{3}^{2} \pmod p$ $\vdots$ $x_{n}^{2}+y_{n}^{2} \equiv x_{1}^{2} \pmod p$

Proof:

We claim that $n=p-1$ satisfies the conditions of the problem. We first consider a system of equations: $x_{1}^{2}+y_{1}^{2} = x_{2}^{2}$ $x_{2}^{2}+y_{2}^{2} = x_{3}^{2}$ $\vdots$ $x_{n}^{2}+y_{n}^{2}=x_{n+1}^{2}$

We repeatedly use the most well-known Pythagorean triple $3^{2}+4^{2} = 5^{2}$ to obtain the following equalities: $(3^{n})^{2}+ (3^{n-1}.4)^{2}=(3^{n-1}.5)^{2}$ $(3^{n-1}.5)^{2}+(3^{n-2}.5.4)^{2}=(3^{n-2}.5^{2})^{2}$ $(3^{n-2}.5^{2})^{2}+(3^{n-3}.5^{2}.4)^{2}=(3^{n-3}.5^{3})^{2}$ $\ldots$ $(3^{n+1-i}.5^{i-1})^{2}+(3^{n-i}.5^{i-1}.4)^{2}=(3^{n-i}.5^{i})^{2}$ $\vdots$ $(3.5^{n-1})^{2}+(5^{n-1}.4)^{2}=(5^{n})^{2}$

Indeed, we set $x_{i} = 3^{n+1-i}.5^{i-1}$, $y_{i}=4.3^{n-i}.5^{i-1}$ for every $i=1, \ldots, n$, and $x_{n+1}=5^{n}$.

To finish our proof, we only need to note that by Fermat’s Little Theorem, we have $x_{2}^{n+1}-x_{1}^{2} \equiv 5^{2n}-3^{2n} \equiv 25^{p-1}-9^{p-1} \equiv 0 \pmod p$,

note that there are infinitely many such n, for instance all multiples of $p-1$.

Ref: 104 Problems Number Theory Problems (from the training of the USA IMO Team) by Prof. Titu Andreescu, Dorin Andrica and Zumin Feng.

More later,

Nalin Pithwa

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