# Can we exploit the unique totient property of primes for a solution to Goldbach's Conjecture (strong)?

The following setup takes advantage of the fact that the totient of every prime p is p-1:

1. Use an “All or nothing” approach in that: $$4\leq 2n=p+q\quad p,q\in\mathbb{P}\iff\forall 2n\geq 4, 2n=r+t\quad r,t\in\mathbb{P}$$

2. Use the totient-quotient value (see below) as a multiplicative identity (Quotient is equal to 1 iff both a and b are prime (non-trivial sum), and not equal to 1 iff both a and b are not prime (trivial sum)).

3. Plot even numbers greater than or equal to 4 with even numbers along the y-axis and sum pairs (not necessarily prime) along the x-axis. The resulting picture is that of an ever-ascending staircase, bridge or chain with each step, section or link, extending further than the last (The number of trivial and non-trivial sums increase as the even numbers increase).

4. Only the quotients that are equal to 1 (non-trivial sums) when multiplied by each “stair step” (the even numbers along the y-axis) fall on that “stair step”, whereas all trivial sums either fall short or surpass it (This is because the totient-quotient value for trivial sums is less or greater than 1). You see, it’s as if the prime sums—the non-trivial sums—are supporting the even numbers as though they provide their structural integrity, not unlike how piers, or supporting columns, keep physical bridges from collapsing under their own weight.

5. Substituting the number 1 with this “re-phrasing” of 1 as a multiplicative identity, in calculating the double factorial of even numbers from some 2n down to 4, e.g., $$(64\times 1)\times(62\times 1)\times(60\times 1)\times\cdots\times (4\times 1) = 64!!$$, we have a contradiction on our hands. If for some even number 2n, there exists no totient-quotient equal to 1, or no non-trivial sum, since, in the case above, 64!! would be less or greater than the actual value of 64!!, which is to say there would be two distinct products for a given set of prime factors, which contradicts the fundamental theorem of arithmetic.

Totient-quotient function: $$\frac{\phi(q)}{(2(n-1)-\phi(p))} =\begin{cases} 1\quad p,q\in\mathbb{P}\\m\neq{1}\quad \text{otherwise}\end{cases} \\2\leq p\leq q,4\leq n, p+q=n$$

• Why is the claim in 1. true? – Wojowu Apr 16 at 20:26
• @Wojowu Proposition 1 is a result of proposition 4 and 5 and the totient function defined. If we were to start at 4 and work our way up to some 2n finding at least one non-trivial sum for each even number only to eventually find an even number with no non-trivial sum, then there would not have existed a non-trivial sum for the preceding even number, and so on, down to 4. The idea is that, primarily, it's not that each individual even number possibly has at least one prime sum, but that prime sums are an inherent property of, that is to say, exist for even numbers as they stand as a collective. – user664788 Apr 16 at 21:18
• please work on your MathJax. – Roddy MacPhee Apr 16 at 21:48
• @Roddy MacPhee Really? Are you trolling me? – user664788 Apr 17 at 0:13
• It's standard precedure on this sight to use MathJax, and show work etc. – Roddy MacPhee Apr 17 at 0:15