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.

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TopNewestNote 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. – Calvin Lin Staff · 2 years, 3 months ago

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– Geerasee Wijesuriya · 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.Log in to reply

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

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– Geerasee Wijesuriya · 2 years, 2 months ago

No. Actually, according to my previous argument, [b- 11^(1/2)].[b+11^(1/2)] / k1 = a. But, if you put a square number(k^2) instead of 11, then gcd(a, b+k) or gcd(a, b-k) not equals to 1. Then according to the gcd operations, gcd(k,a) =1 and gcd(k,b) =1. (The operation is gcd(a,b) = gcd(a+m.b ,b) ). Therefore, if you choose k^2 instead of 11, such that gcd(k,ab) =1.Log in to reply