The sum of two given fractions reduced to their lowest terms is an integer. Their denominators are positive. What is the maximum possible difference between their denominators?

[SOLVED]

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Andrew B. solved this puzzle:

Let the fractions be \(\frac{a}{b}\) and \(\frac{c}{d}\), with \(b \leq d\) without loss of generality, and let their sum be \(n\). Then \(\frac{c}{d} = \frac{nb - a}{b}\), so \(\frac{c}{d}\) can be expressed as a fraction with \(b\) as the denominator. Therefore, \(b - d = 0\).

Hagen von Eitzen solved this puzzle:

The only possible difference is \(0\).

If \(\frac{a}{b} + \frac{c}{d} = n\) and \(\gcd(a, b) = \gcd(c, d) = 1\), then \(\frac{a}{b} = \frac{nd - c}{d}\). If \(e\) is a common divisor of \(nd - c\) and \(d\), then it is also a common divisor of \(c = nd - (nd -c)\) and \(d\). Hence \(\gcd(nd - c, d) = 1\), and by uniqueness of representation of a fraction in lowest terms, \(b = d\).

quasi solved this puzzle:

Suppose \(\frac{a}{b} + \frac{c}{d} = n\), where \(\frac{a}{b}\) and \(\frac{c}{d}\) are reduced fractions, \(b \gt 0\), \(d \gt 0\) and \(n\) is an integer.

Then \(\frac{c}{d} = n - \frac{a}{b}\). Therefore, \(\frac{c}{d} = \frac{bn - a}{b}\). Since \(a\) and \(b\) are relatively prime, \(bn - a\) and \(b\) are also relatively prime. Thus, \(\frac{bn - a}{b}\) is a reduced fraction, so \(\frac{c}{d} = \frac{bn - a}{b}\). It implies that \(c = bn - a\) and \(d = b\). Therefore, \(d - b = 0\).

Thomas Nordhaus solved this puzzle:

Let \(\frac{a}{b} + \frac{c}{d} = n\) with integers \(a\), \(b\), \(c\), \(d\) and \(n\), \(b \gt 0\), \(d \gt 0\), and \(\gcd(a, b) = gcd(c, d) = 1\). Cross-multiplying by \(bd\) results in \(ad + bc = nbd\). This can be written as \(bc = d(nb - a)\) or \(ad = b(nd - c)\). Since the right-hand sides are divisble by d and b, respectively, so are the left-hand sides. Therefore \(d \mid bc\) and \(b \mid ad\). Since \(\gcd(d, c) = 1\) and \(gcd(b,a) = 1\), it follows that \(d \mid b\) and \(b \mid d\). This is possible only if \(b = d\). Therefore, the maximum difference of the denominators is \(0\).

Corey Derochie solved this puzzle:

As per the problem statement, let \(\frac{a}{b} + \frac{c}{d} = n\), where \(a\), \(b\), \(c\), \(d\), and \(n\) are integers. In particular, \(b\) and \(d\) are positive integers. \(a\) and \(b\) are coprime to each other because \(\frac{a}{b}\) is in lowest terms. \(c\) and \(d\) are coprime to each other because \(\frac{c}{d}\) is in lowest terms.

From \(\frac{a}{b} + \frac{c}{d} = n\), we get \begin{equation} \label{eq1} b(nd - c) = ad. \end{equation} \(b\) does not equal \(0\) because \(\frac{a}{b}\) is defined. Suppose \((nd - c) = 0\). Then \(nd = c\). Thus, \(d\) divides \(c\). But \(c\) and \(d\) are coprime to each other, so \(d = 1\). Thus, \(c = n\). Necessarily, \(a = 0\) and \(b = 1\), to satisfy the constraints and \eqref{eq1}. Therefore, \(b\) divides \(d\).

If \((nd - c) \ne 0\), \(\gcd(a, b) = 1\), so \(a\) divides \((nd - c)\) by \eqref{eq1} and unique prime factorization. Thus, for some integer \(m\), \(ma = n*d - c\). Then \(bma = ad\), by \eqref{eq1}. Cancelling \(a\) on both sides we get \(bm = d\). Thus, \(b\) divides \(d\).

Similarly, we can show that \(d\) divides \(b\). Thus, \(b = d\) and \(b - d = 0\).

Siddhesh Chaubal solved this puzzle:

Let the two fractions in reduced form be \(\frac{n_1}{d_1}\) and \(\frac{n_2}{d_2}\). \(\newcommand{\lcm}{\operatorname{lcm}}\)

Let \(\lcm(d_1, d_2) = d\), and let \(d = x_1 d_1 = x_2 d_2\). The sum \(\frac{n_1}{d_1} + \frac{n_2}{d_2}\) is an integer, i.e. \(\frac{n_1 x_1 + n_2 x_2}{d}\) is an integer. Since \(x_1 \mid d\) and \(d \mid (n_1 x_1 + n_2 x_2)\), \(x_1 \mid (n_1 x_1 + n_2 x_2)\). This implies that \(x_1 \mid n_2 x_2\).

\(d = x_1 d_1 = x_2 d_2\) is the LCM of \(d_1\) and \(d_2\). If \(\gcd(x_1, x_2) = g \gt 1\), then \(d_1 \mid \frac{d}{g}\) and \(d_2 \mid \frac{d}{g}\), so \(\lcm(d_1, d_2) \lt d\). This is a contradiction, so \(\gcd(x_1, x_2) = 1\). Therefore, \(x_1 \mid n_2\).

Also, since \(x_1 \mid x_2 d_2\) and \(\gcd(x_1, x_2) = 1\), \(x_1 \mid d_2\). Since, \(\frac{n_2}{d_2}\) is in the lowest form, \(x_1 = 1\). By a similar argument, \(x_2 = 1\). Therefore, \(d = d_1 = d_2\). Thus, the answer is \(0\).

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