**God Has OCD – Part Deux**

Did you know that a shortage of remainders is why the hypotenuse can’t have prime factors in common with the legs of a right triangle? Interesting eh? Want to know more? Then read on.

**Lock The Gates!**

In part 1 of * God Has OCD* we dealt with the Pythagorean Theorem A

^{2}+ B

^{2}= C

^{2}

and we * proved *that, for

*, A and B do not ever “share” prime factors with C. That is, A and B do not ever have a prime factor that is also a prime factor of C.*

**primitive Pythagorean Triples**In this article we’ll look into the “sharing” of prime fctors for the more general ** Sum Of Powers **formulas of which the

*is merely a special case.*

**Pythagorean Theorem**We can express the general *Sum Of Powers* formulas like so:

**T _{1}^{n} + T_{2}^{n} + T_{3}^{n} .. + T_{x}^{n} = Z^{n}**

In the above we’ll call T_{1}^{n} + T_{2}^{n} + T_{3}^{n} .. + T_{x}^{n} the “**terms**” of the equation.

We’ll demonstrate how, for “primitive” N-tuples, where the greatest common denominator of all the terms is 1, that the terms may or may not, “share” prime factors with Z; except for, again, the special case of the *Pythagorean Theorem* where prime factors are never “shared” and where there are only 2 terms (A and B as we’re accustomed to seeing them called).

So where do we begin? As I often do, I experimentally test theories first, and then see if they can be proven. In this case, I could again use my *SumOfPowers* program (in C#).

**With the SumOfPowers program we can specify:**

**How many terms to use****A range of values for the terms****The exponent (n) to use**

The program would then find integer solutions if any exist. Follow the link below for an example screen print of the *SumOfPowers* program after an execution.

For this article I ran the *SumOfPowers* program and the results demonstrated that sometimes the terms T_{1}..T_{x }will have a prime factor in common with Z and sometimes none of the terms will have a prime factor in common with Z.

That said, we now unequivocally know and can demonstrate the answer with a real example but the question is **why?** And then, why is this NOT true for the case of 2 terms (e.g. the *Pythagorean Theorem*)?

Anyway here are a few simple examples found by the *SumOfPowers* program:

**T _{1}^{3} + T_{2}^{3} + T_{3}^{3} .. + T_{6}^{3} = Z^{3}**

1^{3} + 2^{3} + 3^{3} + 6^{3 }+ 16^{3} + 17^{3} = 21^{3}

1^{3} + 2^{3} + 6^{3} + 6^{3 }+ 6^{3} + 7^{3} = 10^{3}

3^{3} + 4^{3} + 5^{3} = 6^{3}

2^{3} + 5^{3} + 12^{3} + 12^{3 }+ 15^{3} + 17^{3} = 24^{3}

**T _{1}^{5} + T_{2}^{5} + T_{3}^{5} .. + T_{8}^{5} = Z^{5}**

8^{5} + 14^{5} + 16^{5} + 20^{5 }+ 20^{5} + 22^{5} + 24^{5} + 30^{5} = 34^{5}

So why is it possible for the *Sum Of Powers* for more than 2 terms to share prime factors with Z while it is NOT possible when there are just 2 terms?

T_{1}^{n} + T_{2}^{n} .. + T_{x}^{n} = Z^{n }** << YES prime factors sometimes shared!**

T_{1}^{n} + T_{2}^{n} = Z^{n}** << NO. Less than 3 terms ****then Prime factors NEVER shared!**

The reason is as follows… If we divide both sides of the equation by one of Z’s prime factors we get something like this (for example):

3^{3} + 4^{3} + 5^{3} = 6^{3} Prime factors of 6 are 3 and 2. Let’s use 3…

3^{3} + 4^{3} + 5^{3} = 6^{3}

—- —— ——- ——–

3 3 3 3

3 divides 3^3 and it also divides 6^3

however (4^3)/3 and (5^3)/3 are not integers but the equation is still valid because

**The sum of all of the remainders **is an integer multiple of 3. For example:

(4^3)/3 = 21 + 1/3 and

(5^3)/3 = 41 + 2/3 then

1/3 + 2/3 = 3/3 = integer

So it all works out because we still get integers on both sides of the equation.

**But note that this could never work out when there’s only 2 terms because then there would only be a single remainder after dividing the terms T**_{1}** and T**_{2}** by one of the prime factors of Z. And with only a single remainder, the left side of the equation can’t possibly be an integer! **

**For example, for A**^{2}** + B**^{2}** = C**^{2 }

let’s assume that one of the prime factors of C was also a prime factor of A. Let’s call that common prime factor P_{c}. Then we would have something like the following if we divide both sides of the equation by P_{c}:

(P_{c} x P_{a2} … x P_{an}) x A + B^{2} = (P_{c} x P_{c2} … x P_{cn}) x C

—————————— ————— ————————————-

P_{c} P_{c} P_{c}

Integer + ( B^{2} / P_{c}) = Integer

But B^{2} / P_{c} yields some remainder after division; And…

**Unlike the case of more than 2 terms on the left side of the equation, **

**there are no additional remainders possible to add to this remainder **

**in order to obtain an integer that would make the equation valid!!!**

Integer + ( B^{2} / P_{c}) = Integer

Integer + NON-Integer = Integer <<=== a contradiction

So our original assumption of some prime factor of C also being a prime factor of A can NOT ever be true for the * Pythagorean Theorem* or for any Sum Of Powers formula with just 2 terms on the left of the equals sign.

**And in some quirky and curious sense **

**we can say that this is because, **

**for just 2 terms (like the Pythagorean Theorem), **

**there will always be a “shortage of remainders”**

**that would prevent the validity of the equation!!!**

All done. We can go now.

**Open The Gates!**