Difference between revisions of "2013 USAJMO Problems/Problem 1"
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We shall prove that such integers do not exist via contradiction. | We shall prove that such integers do not exist via contradiction. | ||
− | Suppose that <math>a^5b + 3 = x^3</math> and <math>ab^5 + 3 = y^3</math> for integers a and b. Rearranging terms gives <math>a^5b = x^3 - 3</math> and <math>ab^5 = y^3 - 3</math>. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = <math>(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}</math> and b = <math>(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}</math>. Consider a prime p in the prime factorization of <math>x^3 - 3</math> and <math>y^3 - 3</math>. If it has power r_1 in <math>x^3 - 3</math> and power r_2 in <math>y^3 - 3</math>, then 5r_1 - r_2 divides 24 and 5r_2 - r_1 also divides 24. Adding and subtracting the divisions gives that r_1 - r_2 divides 12. Because 5r_1 - r_2 also divides 12, 4r_1 divides 12 and thus r_1 divides 3. Repeating this trick for all primes in <math>x^3 - 3</math>, we see that <math>x^3 - 3</math> is a perfect cube, say <math>q^3</math>. Then <math>x^3 - q^3 = 3,</math> and | + | Suppose that <math>a^5b + 3 = x^3</math> and <math>ab^5 + 3 = y^3</math> for integers a and b. Rearranging terms gives <math>a^5b = x^3 - 3</math> and <math>ab^5 = y^3 - 3</math>. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = <math>(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}</math> and b = <math>(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}</math>. Consider a prime p in the prime factorization of <math>x^3 - 3</math> and <math>y^3 - 3</math>. If it has power r_1 in <math>x^3 - 3</math> and power r_2 in <math>y^3 - 3</math>, then 5r_1 - r_2 divides 24 and 5r_2 - r_1 also divides 24. Adding and subtracting the divisions gives that r_1 - r_2 divides 12. Because 5r_1 - r_2 also divides 12, 4r_1 divides 12 and thus r_1 divides 3. Repeating this trick for all primes in <math>x^3 - 3</math>, we see that <math>x^3 - 3</math> is a perfect cube, say <math>q^3</math>. Then <math>x^3 - q^3 = 3,</math> and <math>(x-q)(x^2 + xq + q^2) = 3</math>, so that <math>x - q = 1</math> and <math>x^2 + xq + q^2 = 3</math>. Clearly, this system of equations has no integer solutions for x or q, a contradiction, hence completing the proof. |
Therefore no such integers exist. | Therefore no such integers exist. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:43, 16 March 2014
Problem
Are there integers and
such that
and
are both perfect cubes of integers?
Solution
No, such integers do not exist. This shall be proven by contradiction, by showing that if is a perfect cube then
cannot be.
Remark that perfect cubes are always congruent to ,
, or
modulo
. Therefore, if
, then
.
If , then note that
. (This is because if
then
.) Therefore
and
, contradiction.
Otherwise, either or
. Note that since
is a perfect sixth power, and since neither
nor
contains a factor of
,
. If
, then
Similarly, if
, then
Therefore
, contradiction.
Therefore no such integers exist.
Solution 2
We shall prove that such integers do not exist via contradiction.
Suppose that and
for integers a and b. Rearranging terms gives
and
. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a =
and b =
. Consider a prime p in the prime factorization of
and
. If it has power r_1 in
and power r_2 in
, then 5r_1 - r_2 divides 24 and 5r_2 - r_1 also divides 24. Adding and subtracting the divisions gives that r_1 - r_2 divides 12. Because 5r_1 - r_2 also divides 12, 4r_1 divides 12 and thus r_1 divides 3. Repeating this trick for all primes in
, we see that
is a perfect cube, say
. Then
and
, so that
and
. Clearly, this system of equations has no integer solutions for x or q, a contradiction, hence completing the proof.
Therefore no such integers exist.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.