2005 Canadian MO Problems/Problem 2
Problem
Let be a Pythagorean triple, i.e., a triplet of positive integers with .
- Prove that .
- Prove that there does not exist any integer for which we can find a Pythagorean triple satisfying .
Solution
- We have
By AM-GM, we have
where is a positive real number not equal to one. If , then . Thus and . Therefore,
- Now since , , and are positive integers, is a rational number , where and are positive integers. Now if , where is an integer, then must also be an integer. Thus must be an integer.
Now every pythagorean triple can be written in the form , with and positive integers. Thus one of or must be even. If and are both even, then is even too. Factors of 4 can be cancelled from the numerator and the denominator(since every time one of , , , and increase by a factor of 2, they all increase by a factor of 2) repeatedly until one of , , or is odd, and we can continue from there. Thus the term is odd, and thus is odd. Now and are odd, and is even. Thus is not an integer. Now we have reached a contradiction, and thus there does not exist any integer for which we can find a Pythagorean triple satisfying .
See also
2005 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 3 |