Difference between revisions of "2017 USAMO Problems/Problem 1"
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Let <math>x</math> be odd where <math>x>1</math>. We have <math>x^2-1=(x-1)(x+1),</math> so <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},</math> as desired. | Let <math>x</math> be odd where <math>x>1</math>. We have <math>x^2-1=(x-1)(x+1),</math> so <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},</math> as desired. | ||
+ | |||
+ | ==Solution 4== | ||
+ | I claim that the ordered pair <math>(2^{n} - 1, 2^{n} + 1)</math> satisfies the criteria for all <math>n \geq 2.</math> | ||
+ | Proof: | ||
+ | It is easy to see that the order modulo <math>2^{n+1}</math> of <math>(2^{n} - 1)^k</math> is <math>2, </math> since <math>(2^{n} - 1)^2 = 2^{2n} - 2 \cdot 2^{n} + 1 \equiv 1 \mod 2^{n+1}. </math>, and we can assert and prove similarly for the order modulo <math>2^{n+1}</math> of <math>(2^{n} + 1)^k.</math> Thus, the remainders modulo <math>2^{n+1}</math> that the sequences of powers of <math>2^{n} - 1</math> and <math>2^{n} + 1</math> generate are <math>1</math> for even powers and <math>2^{n} - 1</math> and <math>2^{n} + 1</math> for odd powers, respectively. Since <math>2^{n} + 1</math> and <math>2^{n} - 1</math> are both odd for <math>n \geq 2, </math> <cmath>(2^n + 1)^{2^n - 1} + (2^n - 1)^{2^n + 1} \equiv 0 \mod 2^{n+1}. </cmath> Since there are infinitely many powers of <math>2</math> and since all ordered pairs <math>(2^n - 1, 2^n+1)</math> contain relatively prime integers, we are done. | ||
+ | <math>\boxed{}</math> | ||
+ | |||
+ | -fidgetboss_4000 | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year= 2017|before=First Problem|num-a=2}} | {{USAMO newbox|year= 2017|before=First Problem|num-a=2}} |
Latest revision as of 20:25, 18 June 2020
Problem
Prove that there are infinitely many distinct pairs of relatively prime positive integers and such that is divisible by .
Solution 1
Let . Since , we know . We can rewrite the condition as
Assume is odd. Since we need to prove an infinite number of pairs exist, it suffices to show that infinitely many pairs with odd exist.
Then we have
We know by Euler's theorem that , so if we will have the required condition.
This means . Let where is a prime, . Then , so Note the condition that guarantees that is odd, since
This makes . Now we need to show that and are relatively prime. We see that By the Euclidean Algorithm.
Therefore, for all primes , the pair satisfies the criteria, so infinitely many such pairs exist.
Solution 2
Take . It is obvious (use the Euclidean Algorithm, if you like), that , and that .
Note that
So
Since , all such pairs work, and we are done.
Solution 3
Let be odd where . We have so This means that and since x is odd, or as desired.
Solution 4
I claim that the ordered pair satisfies the criteria for all Proof: It is easy to see that the order modulo of is since , and we can assert and prove similarly for the order modulo of Thus, the remainders modulo that the sequences of powers of and generate are for even powers and and for odd powers, respectively. Since and are both odd for Since there are infinitely many powers of and since all ordered pairs contain relatively prime integers, we are done.
-fidgetboss_4000
See Also
2017 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |