1
$\begingroup$

I'm looking for four prime numbers $a < b < c < z$,

so that $\forall p, q, r, s \in N_{\geq 0}$:

$pa+qb+rc = sz \rightarrow$ "All of $p$, $q$ and $r$ are multiples of $z$"

  1. Do such prime numbers $a, b, c, z$ exist?

For example, $2, 5, 7, 11$ doesn't satisfy the requirements, because $0*2 + 3*5 + 1*7 = 2*11$ and $3$ is not a multiple of $11$ (and by the way neither is $1$ a multiple of $11$). Some other counterexamples for this particular instance are $2*2+0*3+1*7 = 1*11$ and $4*2+1*3+11*7=8*11$

  1. Do we have some examples of such combinations of prime numbers?
  2. Is there an "easy" way to find such combinations?

For my particular application, I don't know yet if I prefer small prime numbers or big ones, but at least it would be good if $z<2^{64}$.

$\endgroup$
1
$\begingroup$

I'm sorry, but there are no such primes.

Since $a$ and $b$ are relatively prime, there exist integers $m, n$ such that $ma + nb = 1$. Choose arbitrary values $s, r$ and let $p = (sz -rc)m, q = (sz -rc)n$. Then $$pa + qb + rc = (sz -rc)(ma + nb) + rc = sz$$

Now, it may be that $p < 0$ or $q < 0$. If $p < 0$, then we can add $taz$ to both sides: $$(p + tz)a + qb + rc = (s + ta)z$$

By choosing $t$ large enough, we get an example with the coefficients of $a$ and $z$ positive. Similarly, adding a multiple of $bz$ to both sizes will result in an example with a positive coefficient for $b$.

Since no requirement was placed on $r$, it can be chosen to not be divisible by $z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.