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.