Difference between revisions of "2013 USAJMO Problems/Problem 1"
(→Solution 3) |
(another solution) |
||
Line 26: | Line 26: | ||
Let <math>a^5b+3=x^3</math> and <math>ab^5+3=y^3</math>. Then, <math>a^5b=x^3-3</math>, <math>ab^5=y^3-3</math>, and <cmath>(ab)^6=(x^3-3)(y^3-3)</cmath> | Let <math>a^5b+3=x^3</math> and <math>ab^5+3=y^3</math>. Then, <math>a^5b=x^3-3</math>, <math>ab^5=y^3-3</math>, and <cmath>(ab)^6=(x^3-3)(y^3-3)</cmath> | ||
Now take <math>\text{mod }9</math> (recall that perfect cubes <math>\equiv -1,0,1\pmod{9}</math> and perfect sixth powers <math>\equiv 0,1\pmod{9}</math>) on both sides. There are <math>3\times 3=9</math> cases to consider on what values <math>\text{mod }9</math> that <math>x^3</math> and <math>y^3</math> take. Checking these <math>9</math> cases, we see that only <math>x^3\equiv y^3\equiv 0\pmod{9}</math> or <math>x\equiv y\equiv 0\pmod{3}</math> yield a valid residue <math>\text{mod }9</math> (specifically, <math>(x^3-3)(y^3-3)\equiv 0\pmod{9}</math>). But this means that <math>3\mid ab</math>, so <math>729\mid (ab)^6</math> so <cmath>729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)</cmath> contradiction. | Now take <math>\text{mod }9</math> (recall that perfect cubes <math>\equiv -1,0,1\pmod{9}</math> and perfect sixth powers <math>\equiv 0,1\pmod{9}</math>) on both sides. There are <math>3\times 3=9</math> cases to consider on what values <math>\text{mod }9</math> that <math>x^3</math> and <math>y^3</math> take. Checking these <math>9</math> cases, we see that only <math>x^3\equiv y^3\equiv 0\pmod{9}</math> or <math>x\equiv y\equiv 0\pmod{3}</math> yield a valid residue <math>\text{mod }9</math> (specifically, <math>(x^3-3)(y^3-3)\equiv 0\pmod{9}</math>). But this means that <math>3\mid ab</math>, so <math>729\mid (ab)^6</math> so <cmath>729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)</cmath> contradiction. | ||
+ | |||
+ | ==Solution 4== | ||
+ | If <math>a^5b+3</math> is a perfect cube, then <math>a^5b</math> can be one of <math>5,6,7 \pmod 9</math>, so <math>(a^5b)^5 = a^{25} b^5</math> can be one of <math>5^5 \equiv 2</math>, <math>6^5 \equiv 0</math>, or <math>7^5 \equiv 4 \pmod 9</math>. If <math>a</math> were divisible by <math>3</math>, we'd have <math>a^5 b \equiv 0 \pmod 9</math>, which we've ruled out. So <math>\gcd(a,9) = 1</math>, which means <math>a^6 \equiv 1 \pmod 9</math>, and therefore <math>a^{25} b^5 \equiv ab^5 \pmod 9</math>. | ||
+ | |||
+ | We've shown that <math>a b^5</math> can be one of <math>0, 2, 4 \pmod 9</math>, so <math>ab^5 + 3</math> can be one of <math>3, 5, 7 \pmod 9</math>. None of these are possibilities for a perfect cube, so if <math>a^5b+3</math> is a perfect cube, <math>ab^5+3</math> cannot be. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:05, 27 July 2017
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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.