Find all quadruples of positive integers d>c>b>a such that 2/a+1/b+1/c1/d=1
holds?
We need 2/n+1/a+1/b1/c=1, with c>b>a>n
n cannot be less than 3, because 2/1 and 2/2 are already too big.
n cannot be more than 3, because then a>=4, and b>=5: 1/4+1/5 = 9/20, which is less than 1/2 (2/4) even ignoring c.
so n=3. Then a cannot be greater than 5, since 1/6+1/7 = 13/42, which is less than 1/3 even ignoring c.
If a=4, then b(c+12) = 12c, with solutions {b,c} = {6,12},{8,24},{9,36},{10,60}{11,132} and we need look no further since 1/4+1/12 = 1/3, even ignoring c.
If a=5, then b(2c+15) = 15c, with solutions {b,c} = {6,30} and {7,105} with no other valid candidates in the positive integers.

Posted by broll
on 20190630 22:03:32 