Question

Let $a,b$ be natural numbers with $ab>2.$ Suppose that the sum of their greatest common divisor and least common multiple is divisible by $a+b.$ Prove that the quotient is at most $\frac{(a+b)}{4}.$ When is this quotient exactly equal to $\frac{(a+b)}{4}?$

Answer 1

Let g and l denote the greatest common divisor and the least common multiple, respectively, of a and b.
Then $gl=ab$.
$\therefore g+l\le ab+1$.
Suppose that $(g+l)=(a+b)>(a+b)/4$.
Then we have $ab+1>(a+b{)}^2/4$, so we get $(a-b{)}^2<4$.
Assuming, $a\ge b$, we either have $a=b$ or $a=b+1$.
In the former case, $g=l=a$ and the quotient is $(g+l)=(a+b)=1\le (a+b)=4$.
In the later case, $g=1$ and $l=b(b+1)$, so we get that $2b+1$ divides $b^2+b+1$.
$\therefore 2b+1$ divides $4(b^2+b+1)-(2b+1{)}^2=3$ which implies that $b=1$ and $a=2$, a contradiction to the given assumption that ab > 2.
This shows that $(g+l)=(a+b)\le (a+b)/4$.
Note that for the equality to hold, we need that either $a=b=2$ or, $(a-b{)}^2/4$ and $g=1,l=ab$.
The later case happens if and only if a and b are two consecutive odd numbers..... (If $a=2k+1$ and $b=2k-1$ then $a+b=4k$ divides $ab+1=4k^2$ and the quotient is precisely $(a+b)/4$.)