Difference between revisions of "2005 USAMO Problems/Problem 2"
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which cannot be true. We now know that | 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 = 3*7^m, 3^m</math>. |
Suppose that | Suppose that | ||
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which is a contradiction. Now suppose that | which is a contradiction. Now suppose that | ||
− | <math>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}- | + | <math>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}-3^{k}+1 = 7^m \rightarrow 3^k(3^{k-1}-1) = 7^m-1</math>. |
We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation to obtain | We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation to obtain |
Revision as of 22:45, 23 March 2012
Contents
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
(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
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 |