Difference between revisions of "Talk:1988 IMO Problems/Problem 6"
(Idea with proof via Chinese Remainder Theorem) |
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I just wonder if it's possible to solve this problem with Chinese Remainder Theorem | I just wonder if it's possible to solve this problem with Chinese Remainder Theorem | ||
| − | First: assuming | + | First: assuming that GCD(a,b)=1. |
| − | Then quotient is always square mod a and mod b and is less or equal than a times b | + | Then quotient is always square mod a and mod b and is less or equal than a times b and is not divisible by neither a nor b which implies it's square of integer. |
In case of GCD(a,b) = d>1 we can transform quotient to d^2((a_1)^2 + (b_1)^2)/(d^2*a_1*b_1 + 1) where a_1 = a/d and b_1 = b/d and follow the same reasoning as above. | In case of GCD(a,b) = d>1 we can transform quotient to d^2((a_1)^2 + (b_1)^2)/(d^2*a_1*b_1 + 1) where a_1 = a/d and b_1 = b/d and follow the same reasoning as above. | ||
| + | |||
| + | It's just an idea without final and rigorous proof. | ||
Am I mistaken? | Am I mistaken? | ||
Help :) | Help :) | ||
Revision as of 07:22, 2 July 2024
I just wonder if it's possible to solve this problem with Chinese Remainder Theorem
First: assuming that GCD(a,b)=1.
Then quotient is always square mod a and mod b and is less or equal than a times b and is not divisible by neither a nor b which implies it's square of integer.
In case of GCD(a,b) = d>1 we can transform quotient to d^2((a_1)^2 + (b_1)^2)/(d^2*a_1*b_1 + 1) where a_1 = a/d and b_1 = b/d and follow the same reasoning as above.
It's just an idea without final and rigorous proof.
Am I mistaken?
Help :)
