Difference between revisions of "2011 USAJMO Problems/Problem 1"
Olympushero (talk | contribs) |
Olympushero (talk | contribs) |
||
Line 47: | Line 47: | ||
We will first take the expression modulo <math>3</math>. We get <math>2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3</math>. | We will first take the expression modulo <math>3</math>. We get <math>2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3</math>. | ||
− | + | Lemma 1: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>. | |
We can prove this by testing the residues modulo <math>3</math>. We have <math>0^2 \equiv 0 \pmod 3</math>, <math>1^2 \equiv 1 \pmod 3</math>, and <math>2^2 \equiv 1 \pmod 3</math>, so the lemma is true. | We can prove this by testing the residues modulo <math>3</math>. We have <math>0^2 \equiv 0 \pmod 3</math>, <math>1^2 \equiv 1 \pmod 3</math>, and <math>2^2 \equiv 1 \pmod 3</math>, so the lemma is true. | ||
Line 54: | Line 54: | ||
Now, we take the original expression modulo <math>4</math>. For right now, we will assume that <math>n>1</math>, and test <math>n=1</math> later. For <math>n>1</math>, <math>2^n \equiv 0 \pmod 4</math>, so <math>2^n+12^n+2011^n=-1^n \pmod 4</math>. | Now, we take the original expression modulo <math>4</math>. For right now, we will assume that <math>n>1</math>, and test <math>n=1</math> later. For <math>n>1</math>, <math>2^n \equiv 0 \pmod 4</math>, so <math>2^n+12^n+2011^n=-1^n \pmod 4</math>. | ||
− | + | Lemma 2: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>. | |
We can prove this by testing the residues modulo <math>4</math>. We have <math>0^2 \equiv 0 \pmod 4</math>, <math>1^2 \equiv 1 \pmod 4</math>, <math>2^2 \equiv 0 \pmod 4</math>, and <math>3^2 \equiv 1 \pmod 4</math>, so the lemma is true. | We can prove this by testing the residues modulo <math>4</math>. We have <math>0^2 \equiv 0 \pmod 4</math>, <math>1^2 \equiv 1 \pmod 4</math>, <math>2^2 \equiv 0 \pmod 4</math>, and <math>3^2 \equiv 1 \pmod 4</math>, so the lemma is true. | ||
Revision as of 13:38, 26 August 2020
Find, with proof, all positive integers for which
is a perfect square.
Solution 1
Let .
Then
.
Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd.
Proof by Contradiction:
We wish to show that the only value of
that satisfies is
.
Assume that
.
Then consider the equation
.
From modulo 2, we easily know x is odd. Let
, where a is an integer.
.
Dividing by 4,
.
Since
,
, so
similarly, the entire LHS is an integer, and so are
and
.
Thus,
must be an integer.
Let
.
Then we have
.
.
.
Thus, n is even.
However, it has already been shown that
must be odd. This is a contradiction. Therefore,
is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.
Solution 2
If , then
, a perfect square.
If is odd, then
.
Since all perfect squares are congruent to , we have that
is not a perfect square for odd
.
If is even, then
.
Since , we have that
is not a perfect square for even
.
Thus, is the only positive integer for which
is a perfect square.
Solution 3
Looking at residues mod 3, we see that must be odd, since even values of
leads to
. Also as shown in solution 2, for
,
must be even. Hence, for
,
can neither be odd nor even. The only possible solution is then
, which indeed works.
Solution 4
Take the whole expression mod 12. Note that the perfect squares can only be of the form 0, 1, 4 or 9 (mod 12). Note that since the problem is asking for positive integers, is always divisible by 12, so this will be disregarded in this process. If
is even, then
and
. Therefore, the sum in the problem is congruent to
, which cannot be a perfect square. Now we check the case for which
is an odd number greater than 1. Then
and
. Therefore, this sum would be congruent to
, which cannot be a perfect square. The only case we have not checked is
. If
, then the sum in the problem is equal to
. Therefore the only possible value of
such that
is a perfect square is
.
Solution 5
We will first take the expression modulo . We get
.
Lemma 1: All perfect squares are equal to or
modulo
.
We can prove this by testing the residues modulo
. We have
,
, and
, so the lemma is true.
We know that if is odd,
, which satisfies the lemma's conditions. However, if
is even, we get
, which does not satisfy the lemma's conditions. So, we can conclude that
is odd.
Now, we take the original expression modulo . For right now, we will assume that
, and test
later. For
,
, so
.
Lemma 2: All perfect squares are equal to or
modulo
.
We can prove this by testing the residues modulo
. We have
,
,
, and
, so the lemma is true.
We know that if is even,
, which satisfies the lemma's conditions. However, if
is odd,
, which does not satisfy the lemma's conditions. Therefore,
must be even.
However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test . We know that
, so the only integer is
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.