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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.

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Comments

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.

Calvin Lin Staff - 6 years, 8 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.

Geerasee Wijesuriya - 6 years, 8 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 $a = 1$ .

Calvin Lin Staff - 6 years, 8 months ago

@Calvin Lin – 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.

Geerasee Wijesuriya - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$