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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>. | ||
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+ | ==Hint== | ||
+ | Consider solving the problem by taking both equations '''mod 19'''. | ||
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== Solution == | == Solution == |
Revision as of 10:26, 19 April 2014
Contents
[hide]Problem
(Răzvan Gelca) Prove that the system has no solutions in integers , , and .
Hint
Consider solving the problem by taking both equations mod 19.
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, 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
(3) , , .
Noting that 73 is a prime number we see that it must divide at least 1 of the 2 factors on the right hand side of 3. Let us consider both cases.
.
However
Thus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case.
.
However
.
It can be seen that 11 and 15 are not perfect cubes from the following.
We can now see that . Furthermore, notice that if
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 when to obtain
.
For we can see that
which is a contradiction. Therefore there only possible solution is when
no integer solutions for k.
Plugging this back into (1) and (2) yields
(4) .
In order for (4) to be true we must have 9 dividing at least 1 of the factors on the right hand side of the equation. Let us consider both cases.
.
However,
.
We now consider the second case.
.
However
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 |
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