Question

(1/x + 1/y) is a factor of​

Answer 1

There are no other solutions besides (2,4,8) and (3,5,15), already discovered in  answer.

Let f(x,y,z)=xyz−1(x−1)(y−1)(z−1) for integers z>y>x>1. We want to know when k=f(x,y,z) is an integer. Note first that k cannot equal 1, because this would imply z=x+y−xyx+y−1 whence xy≤1 which is impossible.

Now

f(x,y,y+1)−f(x,y,z)=(xy−1)(z−(y+1))y(x−1)(y−1)(z−1)≥0(1)

and

f(x,x+1,x+2)−f(x,y,y+1)=t((2x2+2x−1)t+(2x3+4x2−1))(x−1)x(x+1)(y−1)y≥0(2)

where t=y−(x+1), so that f(x,y,z)≤f(x,x+1,x+2)=ϕ(x)=x3+3∗x2+2∗x−1x3−x. If x≥4, we have f(x,x+1,x+2)=11960−(x−4)(100+(x−1)(59x+15))60x(x2−1), so k≤11960<2 and hence k=1 which is impossible as already shown above.

So x can only be 2 or 3. As ϕ(2)=236<4 and ϕ(3)=5924<3, we have k<4 when x=2 and k<3 when x=3.

Note that f(x,y,z)=k can be rewritten as

z=k(x+y)−kxy−k+1k(x+y)−(k−1)xy−k(3)

When x=2 and k=2, (3) yields z=32−y which is impossible.

When x=2 and k=3, (3) yields z=3y−4y−3 which is possible iff y=4.

When x=3 and k=2, (3) yields z=4y−5y−4 which is possible iff y=5.

This concludes the proof.