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1/a+1/b+2007/\lcm (a, b)= 1/\gcd (a, b), 1/a+1/b+2010/\lcm (a, b)= 1/\gcd (a, b)

Source: New Zealand NZMOC Camp Selection Problems 2010 Juniors 4

September 18, 2021
LCMGCDleast common multiplegreatest common divisornumber theory

Problem Statement

Find all positive integer solutions (a,b)(a, b) to the equation 1a+1b+nlcm(a,b)=1gcd(a,b)\frac{1}{a}+\frac{1}{b}+ \frac{n}{lcm(a,b)}=\frac{1}{gcd(a, b)} for (i) n=2007n = 2007; (ii) n=2010n = 2010.
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