Difference between revisions of "2005 USAMO Problems/Problem 2"
(New page: == Problem == (''Răzvan Gelca'') Prove that the system <cmath> \begin{align*} x^6 + x^3 + x^3y + y &= 147^{157} \\ x^3 + x^3y + y^2 + y + z^9 &= 157^{147} \end{align*} </cmath> has no so...) |
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== Problem == | == Problem == | ||
− | + | (''Răzvan Gelca'') Prove that the system | |
− | (''Răzvan Gelca'') | + | <cmath> |
− | Prove that the system | + | \begin{align*}x^6 + x^3 + x^3y + y &= 147^{157} \\ |
− | <cmath> \begin{align*} | + | x^3 + x^3y + y^2 + y + z^9 &= 157^{147}\end{align*} |
− | x^6 + x^3 + x^3y + y &= 147^{157} \\ | + | </cmath> |
− | x^3 + x^3y + y^2 + y + z^9 &= 157^{147} | ||
− | \end{align*} </cmath> | ||
has no solutions in integers <math>x</math>, <math>y</math>, and <math>z</math>. | has no solutions in integers <math>x</math>, <math>y</math>, and <math>z</math>. | ||
== Solution == | == Solution == | ||
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It suffices to show that there are no solutions to this system in the integers mod 19. We note that <math>152 = 8 \cdot 19</math>, so <math>157 \equiv -147 \equiv 5 \pmod{19}</math>. For reference, we construct a table of powers of five: | It suffices to show that there are no solutions to this system in the integers mod 19. We note that <math>152 = 8 \cdot 19</math>, so <math>157 \equiv -147 \equiv 5 \pmod{19}</math>. For reference, we construct a table of powers of five: | ||
<cmath> \begin{array}{c|c||c|c} | <cmath> \begin{array}{c|c||c|c} | ||
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<cmath> (x^3+y+1)^2 \equiv -z^9 - 5 . </cmath> | <cmath> (x^3+y+1)^2 \equiv -z^9 - 5 . </cmath> | ||
By [[Fermat's Little Theorem]], the only possible values of <math>z^9</math> are <math>\pm 1</math> and 0, so the only possible values of <math>(x^3+y+1)^2</math> are <math>-4,-5</math>, and <math>-6</math>. 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. <math>\blacksquare</math> | By [[Fermat's Little Theorem]], the only possible values of <math>z^9</math> are <math>\pm 1</math> and 0, so the only possible values of <math>(x^3+y+1)^2</math> are <math>-4,-5</math>, and <math>-6</math>. 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. <math>\blacksquare</math> | ||
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{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == See also == |
+ | * <url>Forum/viewtopic.php?p=213009#213009 Discussion on AoPS/MathLinks</url> | ||
{{USAMO newbox|year=2005|num-b=1|num-a=3}} | {{USAMO newbox|year=2005|num-b=1|num-a=3}} | ||
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[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Revision as of 12:05, 3 May 2008
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. ! Misplaced alignment tab character &.)
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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 |