Difference between revisions of "2005 USAMO Problems/Problem 2"
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<math>x^2 \ge 2x</math> when <math>|x| \ge 2 \rightarrow x^2-x+1 \ge x+1</math> | <math>x^2 \ge 2x</math> when <math>|x| \ge 2 \rightarrow x^2-x+1 \ge x+1</math> | ||
− | when <math>|x| \ge 2</math>. If <math>x = 1</math> or <math>x = -1</math> then examining (1) would yield <math>147^{157} \equiv 0 \pmod{2}</math> which is a contradiction. | + | when <math>|x| \ge 2</math>. If <math>x = 1</math> or <math>x = -1</math> then examining (1) would yield <math>147^{157} \equiv 0 \pmod{2}</math> which is a contradiction. If <math>x = 0</math> then from (1) we can see that <math>y+1 = 147^{157}</math>, plugging this into 2 yields <math>(147^{157}-1)(147^{157}) = (157^{49}-z^3)(157^{98}+157^{49}z^3+z^6)</math>. We can now see that <math>|x| \ge 2</math>. Furthermore, notice that |
<math>x+1 = 3^k7^j</math> | <math>x+1 = 3^k7^j</math> | ||
− | + | for a pair of positive integers <math>(j,k)</math> means that | |
<math>x^2-x+1 \le 3</math> | <math>x^2-x+1 \le 3</math> | ||
− | + | which cannot be true. We now know that | |
<math>x+1 = 3^k, 7^k \rightarrow x^2-x+1 = 7^m, 3^m</math>. | <math>x+1 = 3^k, 7^k \rightarrow x^2-x+1 = 7^m, 3^m</math>. | ||
− | + | Suppose that | |
<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> | <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> |
Revision as of 21:20, 23 March 2012
Contents
[hide]Problem
(Răzvan Gelca) Prove that the system has no solutions in integers , , and .
Solution
It suffices to show that there are no solutions to this system in the integers mod 19. We note that , so . For reference, we construct a table of powers of five: 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 , and . Thus we rewrite our system thus: Adding these, we have
\[(x^3+y+1)^2 - 1 + z^9 &\equiv -6,\] (Error compiling LaTeX. Unknown error_msg)
or By Fermat's Little Theorem, the only possible values of are and 0, so the only possible values of are , and . 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.
Solution 2
Note that the given can be rewritten as
(1) ,
(2) .
We can also see that
.
Now we notice
for some pair of non-negative integers . We also note that
when
when . If or then examining (1) would yield which is a contradiction. If then from (1) we can see that , plugging this into 2 yields . We can now see that . Furthermore, notice that
for a pair of positive integers means that
which cannot be true. We now know that
.
Suppose that
which is a contradiction. Now suppose that
.
We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation to obtain
.
For we can see that which is a contradiction. Therefore there are no solutions to the given system of diophantine equations.
See also
- <url>Forum/viewtopic.php?p=213009#213009 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |