Difference between revisions of "2013 USAJMO Problems/Problem 1"
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Therefore no such integers exist. | Therefore no such integers exist. | ||
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+ | Amkan2022 | ||
==Solution 2== | ==Solution 2== | ||
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Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes. | Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes. | ||
− | '''Lemma 1''': If <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes <math>ab^5,a^5b \equiv 5,7 \pmod 9</math> | + | '''Lemma 1''': If <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes, then <math>ab^5, a^5b \equiv 5,7 \pmod 9</math> |
'''Proof''' 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>. | '''Proof''' 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>. | ||
− | '''Lemma 2''': If k is a perfect 6th power, <math>k \equiv 0,1 \pmod 9</math> | + | '''Lemma 2''': If k is a perfect 6th power, then <math>k \equiv 0,1 \pmod 9</math> |
+ | |||
+ | '''Proof''': Since cubes are congruent to <math>0, 1, -1 \pmod 9</math>, we can square, and get 6th powers are congruent to <math>0, 1 \pmod 9</math>. | ||
− | |||
Since <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, which is a perfect 6th power, by lemma 2, <math>ab^5 \cdot a^5b \equiv 0,1 \pmod 9</math>. | Since <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, which is a perfect 6th power, by lemma 2, <math>ab^5 \cdot a^5b \equiv 0,1 \pmod 9</math>. | ||
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-AlexLikeMath | -AlexLikeMath | ||
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+ | |||
+ | |||
+ | ==Solution 7== | ||
+ | Note that <math>x^3 \equiv 0,1,-1 \pmod 9</math>. So <math>a^5b+3, ab^5+3\equiv 0,1,-1\pmod{9}</math><math>\Rightarrow a^5b, ab^5 \equiv 5,6,7 \pmod 9</math>. | ||
+ | <math>a^5b \equiv 5 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 5 \pmod 9</math> | ||
+ | <math>a^5b \equiv 6 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 6 \pmod 9</math> | ||
+ | <math>a^5b \equiv 7 \pmod 9 \;\;\;\;\;\; ab^5 \equiv 7 \pmod 9</math>. | ||
+ | Multiplying <math>a^5b</math> and <math>ab^5</math> we get <math>(ab)^6</math>. By substituting values for <math>a^5b</math> and <math>ab^5</math> we can see that <math>(ab)^6 \equiv 0,1,3,6,7,8 \pmod 9</math> but because cubes are not <math>3,6,7 \pmod 9</math>, and squares are not <math>-1\pmod 9</math> so <math>\boxed{(ab)^6 \equiv 0,1 \pmod 9}</math>. | ||
+ | *Case I: <math>(ab)^6\equiv 0 \pmod 9</math>.This is only true when <math>3|ab</math>. If <math>3|a,b</math> then <math>a^5b+3 \equiv 3 \pmod 9</math>, which is not a perfect cube. The same can be seen with if <math>3|a</math> then <math>a^5b+3 \equiv 3 \pmod 9</math> and if <math>3|a</math> then <math>b^5a+3 \equiv 3 \pmod 9</math>. So 6 mod 9 case is eliminated from the above relations of congruences | ||
+ | *Case II: <math>(ab)^6\equiv 1 \pmod 9</math> for this we'll take a look at the above information regarding <math>a^5b</math> and <math>ab^5</math>. If <math>a^5b \equiv 5 \pmod 9</math> then, | ||
+ | <math>(ab)^6=a^5b \cdot ab^5 \equiv 5(ab^5) \pmod 9 \Rightarrow ab^5 \equiv 2 \pmod 9 \Rightarrow ab^5+3 \equiv | ||
+ | 5 \pmod 9</math> | ||
+ | which is not possible. Similarly, if <math>a^5b \equiv 7 \pmod 9</math> then, | ||
+ | <math>(ab)^6=a^5b \cdot ab^5 \equiv 7(ab^5) \pmod 9 \Rightarrow ab^5 \equiv 4 \pmod 9 \Rightarrow ab^5+3 \equiv | ||
+ | 7 \pmod 9</math> | ||
+ | which is not possible. So, no such integers. | ||
+ | |||
+ | -Lakshya Pamecha | ||
+ | |||
+ | ==Note== | ||
+ | After you get that <math>a^5 b \equiv 5, 6, \text{or} 7 \pmod{9}</math>, you can also bash by listing all possible residues for <math>a^5</math> and their corresponding values for <math>b</math>. At that point, you can easily see that all the solutions for <math>a</math> and <math>b</math> do not work in the other equation, meaning that there are no solutions that satisfy both equations. | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 02:45, 10 May 2024
Contents
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.
Amkan2022
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
I claim there are no such a or b such that both expressions are cubes.
Assume to the contrary and
are cubes.
Lemma 1: If and
are cubes, then
Proof Since cubes are congruent to any of ,
. But if
,
, so
, contradiction. A similar argument can be made for
.
Lemma 2: If k is a perfect 6th power, then
Proof: Since cubes are congruent to , we can square, and get 6th powers are congruent to
.
Since , which is a perfect 6th power, by lemma 2,
.
But, by lemma 1, .
So, , which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete.
-AlexLikeMath
Solution 7
Note that . So
.
![]()
![]()
.
Multiplying and
we get
. By substituting values for
and
we can see that
but because cubes are not
, and squares are not
so
.
- Case I:
.This is only true when
. If
then
, which is not a perfect cube. The same can be seen with if
then
and if
then
. So 6 mod 9 case is eliminated from the above relations of congruences
- Case II:
for this we'll take a look at the above information regarding
and
. If
then,
which is not possible. Similarly, if then,
which is not possible. So, no such integers.
-Lakshya Pamecha
Note
After you get that , you can also bash by listing all possible residues for
and their corresponding values for
. At that point, you can easily see that all the solutions for
and
do not work in the other equation, meaning that there are no solutions that satisfy both equations.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.