No.1
若\(a+b+gcd(a,b)=lcm(a,b)\)且\(a<b\),则\(2b=3a\)
证明:由于\(gcd(a,b)\times lcm(a,b)=a\times b\),所以原式可以变为\(gcd(a,b)\times a+gcd(a,b)\times b+gcd(a,b)^{2}=a\times b\)
令\(a=gcd(a,b)\times k_{1}\),\(b=gcd(a,b)\times k_{2}\)(\(k_{1}<k_{2}\))
原式就可以转化为\(k_{1}+ k_{2}+1=k_{1}\times k_{2}\)
移项得到\((k_{1}-1)\times (k_{2}-1)=2\)
由于都为整数且\(k_{1}<k_{2}\)
所以\(k_{1}=2,k_{2}=3\)
得证