Difference between revisions of "2005 USAMO Problems/Problem 2"

Revision as of 19:10, 23 March 2012

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. $\blacksquare$

Solution 2

Note that the given can be rewritten as

$(x^3+y)(x^3+1) = (x^3+y)(x+1)(x^2-x+1) = 147^{157} = 7^{314}3^{152}$ ,

$(x^3+y)(y+1)+z^9 = 157^{147}$ .

We can also see that

$x^2-x+1 = (x+1)^2-3(x+1)+3 \rightarrow gcd(x+1, x^2-x+1) \le 3$ .

Now we notice

$x^3+1 = 3^a7^b$

for some pair of non-negative integers $(a,b)$ . We also note that

$x^2 \ge 2x$ when $|x| \ge 2 \rightarrow x^2-x+1 \ge x+1$

when $|x| \ge 2$ . Furthermore, notice that

$x+1 = 3^k7^j$

for a pair of positive integers  means that

$x^2-x+1 \le 3$

which cannot be true. We now know that

$x+1 = 3^k, 7^k \rightarrow x^2-x+1 = 7^m, 3^m$ .

Suppose that

$x+1 = 7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}-3*7^k+3 = 3^m \rightarrow 7^{2k} \equiv 0 \pmod{3}$

which is a contradiction. Now suppose that

$x+1 = 3^k \rightarrow (x+1)^2-3(x+1)+3 = 3*7^m \rightarrow 3^{2k}-3^{k+1}+3 = 3*7^m \rightarrow 3^{2k-1}-x*3^{k}+1 = 7^m \rightarrow 3^k(3^{k-1}-1) = 7^m-1$ .

We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation to obtain

$k = 1+v_3(m) \le 1+log_3(m) \rightarrow 7^m-1 = 3^k(3^{k-1}-1)\le 3^{1+log_3(m)}(3^{log_3(m)}-1) = 3m(m-1)$ .

We can see that 7 must divide m and m-1 which cannot be true as they are relatively prime leading us to conclude that there are no solutions to the given system of diophantine equations.

@@ Line 64: / Line 64: @@
   Suppose that
-<math>x+1 = 7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}-37^k+3  = 3^m \rightarrow 7^2k \equiv 0 \pmod{3}</math>
+<math>x+1 = 7^k \rightarrow (x+1)^2-3(x+1)+3 = 3^m \rightarrow 7^{2k}-3*7^k+3  = 3^m \rightarrow 7^{2k} \equiv 0 \pmod{3}</math>
 which is a contradiction. Now suppose that

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Difference between revisions of "2005 USAMO Problems/Problem 2"

Revision as of 19:10, 23 March 2012

Contents

Problem

Solution

Solution 2

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