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For all integers n > 3 there exists two integers 'a' and 'b' such that: n = a(Biggest prime smaller than n) + b(Second biggest prime smaller than n)

Formal Logic Statement:

∀n ∈ {x ∈ ℤ : x > 3} ∃{a; b} ⊂ ℤ | n = a * ⊔P + b * ⊔S where P = {p ∈ ℤ+ : p is prime ∧ p < n} ∧ S = {x ∈ P : x < ⊔P}

Found lots of examples for the first 2000 'n' using brute force. I came across a somewhat similar question about if the sum of two integer multiples of primes can equal any other integer, which could be proven using Bézout's identity.

Any ideas?

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    $\begingroup$ This is just Bezout applied to the two largest primes $<n$. Clearly these are relatively prime (as they are distinct primes). $\endgroup$
    – lulu
    Commented Jun 7, 2023 at 23:58
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Commented Jun 8, 2023 at 0:20
  • $\begingroup$ @lulu so according to what I understand of Bezout's Identity that means that the best we can get is: a*(P1) + b*(P2) = 1. Did I understand that correctly? Considering that there are lots of examples for n<2000 I'm pretty sure I didn't quite understand how Bezout applies here. Mind explaining briefly? $\endgroup$
    – hefe
    Commented Jun 8, 2023 at 15:18
  • $\begingroup$ If you can solve $an+bm=1$ then you can solve $An+Bm=k$ for any $k$. Just let $A=k\times a, B=k\times b$ where $a,b$ solve $an+bm=1$. $\endgroup$
    – lulu
    Commented Jun 8, 2023 at 18:50

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