Therefore, 1/8 can be expressed as the sum of the reciprocal of three mutually different positive integers (the expression may not be unique).
(2) Solution:
We might as well set up a<b<C. Now the problem is to find the whole A, B and C and satisfy1/8 =1/A2+1/B2+1/C2.
By a<b<c, with1/A2 > 1/b2 > 1/c2, so1/8 =1/A2+1/B2+1/C2.
So A2: 8, so a2=9 or 16.
If a2=9, then1/B2+1/C2 =1/8-1/9 =1/72, because1/72
So b2 & gt72 <; b2 & lt 144。
So b2=8 1, 100 or 12 1. Substituting b2=8 1, 100, 12 1 into c2=72b/b2-72, respectively, none of them is a complete square number, indicating that there is no solution when a2=9.
If a2= 16, then1/B2+1/C2 =1/8-1/6 =1/6. Similarly, we can get: 16.
I.e. b2=25,
At this time, C2 =16b2/b2-16 =16 * 25/9 is not an integer.
To sum up, 1/8 cannot be expressed as the sum of the reciprocal of three different complete squares.
Supplementary question:
Let P=7a+2b+3c Q=5a+7b-22c.
3P+Q = 26a+ 13 b- 13c = 13(2a+b-c)| 13
Because P| 13, Q| 13, that is, 5a+7b-22c, can definitely be divisible by 13.