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Calvin Lin:

Find the number of coprime pairs (a, b) of positive integers subject to 0 100, such that a|(b2-11) and b|(a2-11).

There would be a small correction to the problem. You say, a|b2-11. Then a = [ b^2 -11] / k1. Where k1 is a natural number. Then, [b- 11^(1/2)].[b+11^(1/2)] / k1 = a = prime number. Then either [b- 11^(1/2)] /k1 OR [b+11^(1/2)]/k1. Because a>1 according to the given conditions. But b is a natural number. And 11^(1/2) is not a natural number. Also k1 is a natural number. Then we have a contradition. If you input some square number instead of 11, then the problem may okay.

Note by Geerasee Wijesuriya
2 years, 3 months ago

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Note that "coprime pairs of positive integers" means that $$\gcd (a,b) = 1$$.

It does not imply that $$a$$ and $$b$$ are prime numbers.

For example, $$(6, 35 )$$ are coprime, but neither of the numbers are prime. Staff · 2 years, 3 months ago

Calvin Lin: Actually, a and b are not needed to be prime numbers. According to my previous argument, then a =1. Then according to the problem's conditions b can take only 1, 2,5. Then answer should be 3. · 2 years, 3 months ago

Sorry but I do not understand your argument.

Note that $$(5, 14)$$ is also a solution, since $$5^2 - 11 = 14$$ and $$14^2 - 11 = 185$$. This contradicts your claim that $$a = 1$$. Staff · 2 years, 3 months ago