2005 USAMO Problems/Problem 2
(Răzvan Gelca) Prove that the system has no solutions in integers , , and .
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 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.
Note that the given can be rewritten as
We can also see that
Now we notice
for some pair of non-negative integers . We also note that
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.
Thus we can see that 73 cannot divide the first factor in the right hand side of (3). Let us consider the next case.
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
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
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.
We now consider the second case.
Therefore there are no solutions to the given system of diophantine equations.
We will show there is no solution to the system modulo 13. Add the two equations and add 1 to obtain By Fermat's Theorem, when is not a multiple of 13. Hence we compute and . Thus The cubes mod 13 are , , and . Writing the first equation as we see that there is no solution in case and for congruent to , correspondingly must be congruent to . Hence Also is a cube, hence must be . It is easy to check that is not obtained by adding one of to one of . Hence the system has no solutions in integers.
Note: This argument shows there is no solution even if is replaced by .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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