Contraharmonic mean given harmonic mean

Question

Given that two positive integers, $X$ and $Y$, have a harmonic mean of $6.875$, what is their contraharmonic mean. Harmonic mean is $(2XY)/(X+Y)$ and contraharmonic mean is $(X^2 +Y^2)/(X+Y)$. I began by representing the harmonic mean as $55/8$, then solving $2XY=55$ and $X+Y=8$. This set of equations does not have real answers, but $X$ and $Y$ are $5$ and $11$ (which is correct) when $55/8$ is changed to $110/16$ in the solution. $55/8 = 110/16$, so why would the answers be different when you change this ratio? I've also noticed that we are given $(X+Y)$ and $2XY$ which allows us to solve for $X^2 +Y^2$ because $(X+Y)^2-2XY=X^2+Y^2$. The same issue arises in this case, assuming $110/16$ rather then $55/8$ produces the correct answer. Can anyone explain what I am not seeing here? Any help would be greatly appreciated.

Answer 1

Without knowing that $X$ and $Y$ are integers, one cannot determine their contraharmonic mean from their harmonic mean alone. If we denote the harmonic mean by $C$ and the harmonic mean by $H$, note that

$$ C+H = \frac{X^2+2XY+Y^2}{X+Y} = X+Y $$

Therefore, knowing $H$ alone cannot determine $C$. We must use the fact that both numbers are integers. Algebraic manipulation yields

$$ \frac{2XY}{X+Y} = \frac{55}{8} $$ $$ 16XY = 55X+55Y $$ $$ 16XY-55Y = 55X $$ $$ Y = \frac{55X}{16X-55} $$

Without loss of generality, let $X < Y$, then $0 < X < 55/8$. Simple trial and error shows that $X$ can only be $5$, yielding $Y = 11$.