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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 $z<2^{64}$.

Albert Hendriks
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1 Answers1

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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 $z$.

Paul Sinclair
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