Difference between revisions of "Talk:1988 IMO Problems/Problem 6"
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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 that GCD(a,b)=1. | + | First: assuming that <math>GCD(a,b)=1</math>. |
| − | Then quotient is always square mod a and mod b and is less or equal than | + | Then quotient is always square <math>mod a</math> and <math>mod b</math> and is less or equal than <math>ab</math> and is not divisible by neither <math>a</math> nor <math>b</math> 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^ | + | In case of <math>GCD(a,b) = d>1</math> we can transform quotient to <math>d^2((a_1)^2 + (b_1)^2)/(d^2a_1b_1 + 1)</math> where <math>a_1 = a/d</math> and <math>b_1 = b/d</math> and follow the same reasoning as above. |
| − | It's just an idea without final and rigorous proof. | + | It's just an idea without final and rigorous proof yet and it may contain counterexample gaps. |
Am I mistaken? | Am I mistaken? | ||
Help :) | Help :) | ||
Latest revision as of 08:55, 2 July 2024
I just wonder if it's possible to solve this problem with Chinese Remainder Theorem
First: assuming that 
.
Then quotient is always square 
and 
and is less or equal than 
and is not divisible by neither 
nor 
which implies it's square of integer.
In case of 
we can transform quotient to 
where 
and 
and follow the same reasoning as above.
It's just an idea without final and rigorous proof yet and it may contain counterexample gaps.
Am I mistaken?
Help :)
