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How does one get the mandatory relations of a set of nonzero unequal numbers $m_a$ and $n_a$ and nonzero number $b_a$ that satisfies the following relation?

$m_an_a = b_a(m_a+n_a)$

($m_1n_1 = b_1(m_1 + n_1) = m_2n_2 = b_2(m_2 + n_2)....$ where $a represents the cardinality of triplet.)

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One standard approach to this kind of problem is to note that the equation $xy=b(x+y)$ is equivalent to $(x-b)(y-b)=b^2.$ So we are studying the integer solutions of $st=b^2$. These are not hard to describe. Let $s$ be any factor of $b^2$ (possibly negative), and let $t=\dfrac{b^2}{s}$.

So for any fixed $b$, the number of solutions $(x,y)$ is just the number of factors of $b^2$, including the negative ones. There is a straightforward formula for the number of factors in terms of the prime power factorization of $b$. If $|b|=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k},$ where the $p_i$ are distinct primes, then $b^2$ has $2(2e_1+1)(2e_2+1)\cdots(2e_k+1)$ factors, including the negative ones.