How does this program work ?

Let's take the case of (6,2,5).
We need to solve the equation:

For 6th power, we have the relations:

These relations are very important.

Now, we can deduce some restrictions on a, b, c, d, e, f and g !

The first one is that a and b cannot be both multiple of 2, 3 or 7.
If a and b were both multiple of 2 for example, then c, d, e, f and g are necessarily multiple of 2, so all terms are multiple of 2 and the solution is not primitive.

We will try to find c and d that can be related to a and b.
Let's enumerate the conditions on a and b modulo 7.

1) if a⁶ = 0 mod 7 and b⁶ = 0 mod 7 then c⁶ = 0 mod 7, d⁶ = 0 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶= 0 mod 7, so all values are divisible by 7, SOLUTION NOT PRIMITIVE, so IMPOSSIBLE CASE !

2) if a⁶ = 0 mod 7 and b⁶ = 1 mod 7 then:
    c⁶ = 0 mod 7 and d⁶= 1 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶ = 0 mod 7
or:
    c⁶ = 1 mod 7 and d⁶= 0 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶ = 0 mod 7

3) if a⁶ = 1 mod 7 and b⁶ = 0 mod 7 then:
    c⁶ = 0 mod 7 and d⁶= 1 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶ = 0 mod 7
or:
    c⁶ = 1 mod 7 and d⁶= 0 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶ = 0 mod 7

4) if a⁶ = 1 mod 7 and b⁶ = 1 mod 7 then c⁶ = 1 mod 7 and d⁶ = 1 mod 7, e⁶ = 0 mod 7, f⁶ = 0 mod 7, g⁶= 0 mod 7

And we have now:

There is a wonderful trick that can be used there !!!
Suppose that we have a, b and c then we can deduce d because a⁶+b⁶-c⁶-d⁶ is divisible by 7⁶
This is done by using a table of residues modulo 7⁶ =117649.

More informations coming soon.

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