Harristotelian Logic

Consider

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34)) = 2(sqrt(13) - sqrt(6))(sqrt(13) + sqrt(41))

and the proof that (sqrt(13) - sqrt(6)) and (sqrt(13) + sqrt(41)) must share the same factors in common with 7.

That proof is what I outlined before with my first attempt at this kind of demonstration where I finally realized that 2 was not the prime I wanted, as that demonstration failed.

So I went to a higher prime, and choosing 7, I don't run into the same problems I ran into with 2.

As a recap, notice that with

(sqrt(13) - sqrt(6) + sqrt(41) + sqrt(34))

you can also group that as

(sqrt(13) + sqrt(41) - sqrt(6) + sqrt(34))

and in each case only factors of 7 shared between sqrt(13) - sqrt(6) and sqrt(13) + sqrt(41) can be factors of the full expression, which is also true for

(sqrt(13) - sqrt(6) + sqrt(41) - sqrt(34))

and with the right side being a product of sqrt(13) - sqrt(6) and sqrt(13) + sqrt(41), you know they must have the same factors of 7 in common.

Note that since

(sqrt(13) - sqrt(6))*(sqrt(13) - sqrt(6)) = 7

and

(sqrt(13) + sqrt(41))*(sqrt(13) - sqrt(41)) = -28

the first should be a factor of the other, so letting

z = (sqrt(13) + sqrt(41))/(sqrt(13) - sqrt(6))

I find that is a root of

7z^4 - 52z^3 - 100z^2 + 208z + 112 = 0.

So the problem with my previous demonstration was using 2. Here a mix of radicals and the same basic approach work quite well with 7 being the key prime.