2005 USAMO Problems/Problem 2

Problem

(Răzvan Gelca) Prove that the system $\begin{align*}x^6 + x^3 + x^3y + y &= 147^{157} \\ x^3 + x^3y + y^2 + y + z^9 &= 157^{147}\end{align*}$ has no solutions in integers $x$ , $y$ , and $z$ .

Solution

It suffices to show that there are no solutions to this system in the integers mod 19. We note that $152 = 8 \cdot 19$ , so $157 \equiv -147 \equiv 5 \pmod{19}$ . For reference, we construct a table of powers of five: $\begin{array}{c|c||c|c} n& 5^n &n & 5^n \\\hline 1 & 5 & 6 & 7 \\ 2 & 6 & 7 & -3 \\ 3 & -8 & 8 & 4 \\ 4 & -2 & 9 & 1 \\ 5 & 9 && \end{array}$ Evidently, then the order of 5 is 9. Hence 5 is the square of a multiplicative generator of the nonzero integers mod 19, so this table shows all nonzero squares mod 19, as well.

It follows that $147^{157} \equiv (-5)^{13} \equiv -5^4 \equiv 2$ , and $157^{147} \equiv 5^3 \equiv -8$ . Thus we rewrite our system thus: $\begin{align*} (x^3+y)(x^3+1) &\equiv 2 \\ (x^3+y)(y+1) + z^9 &\equiv -8 . \end{align*}$ Adding these, we have

\[(x^3+y+1)^2 - 1 + z^9 &\equiv -6,\] (Error compiling LaTeX. Unknown error_msg)

or $(x^3+y+1)^2 \equiv -z^9 - 5 .$ By Fermat's Little Theorem, the only possible values of $z^9$ are $\pm 1$ and 0, so the only possible values of $(x^3+y+1)^2$ are $-4,-5$ , and $-6$ . But none of these are squares mod 19, a contradiction. Therefore the system has no solutions in the integers mod 19. Therefore the solution has no equation in the integers. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2005 USAMO (Problems • Resources)
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2005 USAMO Problems/Problem 2

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