Difference between revisions of "2013 USAJMO Problems/Problem 1"
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As in previous solutions, notice <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. Now multiplying gives <math>a^6b^6</math>, which is only <math>0,1 \pmod 9</math>, so after testing all cases we find that <math>ab^5\equiv a^5b \equiv 6 \mod 9</math>. Then since <math>\phi (9) = 6</math>, <math>ab^5\equiv \frac{a}{b}\pmod 9</math> and <math>a^5b \equiv \frac{b}{a}\pmod 9</math> (Note that <math>a,b</math> cannot be <math>0\pmod 9</math>). Thus we find that the inverse of <math>6</math> is itself under modulo <math>9</math>, a contradiction. | As in previous solutions, notice <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. Now multiplying gives <math>a^6b^6</math>, which is only <math>0,1 \pmod 9</math>, so after testing all cases we find that <math>ab^5\equiv a^5b \equiv 6 \mod 9</math>. Then since <math>\phi (9) = 6</math>, <math>ab^5\equiv \frac{a}{b}\pmod 9</math> and <math>a^5b \equiv \frac{b}{a}\pmod 9</math> (Note that <math>a,b</math> cannot be <math>0\pmod 9</math>). Thus we find that the inverse of <math>6</math> is itself under modulo <math>9</math>, a contradiction. | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes. Since cubes are congruent to any of <math>0, 1, -1 \pmod 9</math>, <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. But if <math>ab^5 \equiv 6 \pmod 9</math>, <math>3|a</math>, so <math>a^5b \equiv 0 \pmod 9</math>, contradiction. A similar argument can be made for <math>ab^5 \neq 6 \pmod 9</math>. Thus, we get that: | ||
+ | |||
+ | <cmath>ab^5,a^5b \equiv 5,7 \pmod 9</cmath> | ||
+ | |||
+ | This will be important later on. | ||
+ | |||
+ | A critical idea is that since cubes are congruent to <math>0, 1, -1 \pmod 9</math>, 6th powers are congruent to <math>0, 1 \pmod 9</math>. And, <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, making the the product of <math>ab^5, a^5b</math> a perfect 6th power. | ||
+ | |||
+ | But, we can substitiute our possible values of <math>ab^5, a^5b</math>. <math>5 \cdot 5 \equiv 7 \pmod 9</math>, <math>5 \cdot 7 \equiv 8 \pmod 9</math>, and <math>7 \cdot 7 \equiv 4 \pmod 9</math>. None of these are congruent to <math>0, 1 \pmod 9</math>, but perfect 6th powers have to be congruent to <math>0, 1 \pmod 9</math>. We have a contradiction. <math>\blacksquare</math> | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:46, 15 June 2020
Contents
[hide]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 x and y. 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
in
and power
in
, then
-
is a multiple of 24 and
-
also is a multiple of 24.
Adding and subtracting the divisions gives that -
divides 12. (actually,
is a multiple of 4, as you can verify if
. So the rest of the proof is invalid.) Because
-
also divides 12,
divides 12 and thus
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
or
, a contradiction, hence completing the proof.
Therefore no such integers exist.
Solution 3
Let and
. Then,
,
, and
Now take
(recall that perfect cubes
and perfect sixth powers
) on both sides. There are
cases to consider on what values
that
and
take. Checking these
cases, we see that only
or
yield a valid residue
(specifically,
). But this means that
, so
so
contradiction.
Solution 4
If is a perfect cube, then
can be one of
, so
can be one of
,
, or
. If
were divisible by
, we'd have
, which we've ruled out. So
, which means
, and therefore
.
We've shown that can be one of
, so
can be one of
. None of these are possibilities for a perfect cube, so if
is a perfect cube,
cannot be.
Solution 5
As in previous solutions, notice . Now multiplying gives
, which is only
, so after testing all cases we find that
. Then since
,
and
(Note that
cannot be
). Thus we find that the inverse of
is itself under modulo
, a contradiction.
Solution 6
Assume to the contrary and
are cubes. Since cubes are congruent to any of
,
. But if
,
, so
, contradiction. A similar argument can be made for
. Thus, we get that:
This will be important later on.
A critical idea is that since cubes are congruent to , 6th powers are congruent to
. And,
, making the the product of
a perfect 6th power.
But, we can substitiute our possible values of .
,
, and
. None of these are congruent to
, but perfect 6th powers have to be congruent to
. We have a contradiction.
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