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Given any fraction $\frac{s}{t}=\frac{s}{\Pi_pp^{i_p}}$ with $s,t$ relatively prime, I would like to know if it is possible to write $\frac{s}{\Pi_pp^{i_p}}=\sum_p\frac{s_p}{p^{i_p}}$ for some unique integers $s_p$.

It seems like this is a simple algebraic fact, but I'm not totally sure how to prove it, especially the uniqueness part. The furthest that I've gotten is that for all nonzero $i_p$ we must have $s_p$ nonzero, because otherwise $s$ would be divisible by that $p$, contradicting that $t$ and $s$ are relatively prime.

I guess an equivalent question would be to ask for solutions to some finite equation $\sum_i a_ix_i=a$ where $\text{gcd}(a,\{a_i\})=1$.

Miles Johnson
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  • What do you mean by "unique"? If you mean that $s_p\neq s_q$ when $p\neq q$, then it's false. You have things like $\frac56=\frac12+\frac13$, with $s_2=s_3=1$. – 2'5 9'2 Nov 21 '19 at 04:41
  • @alex.jordan I mean like if $5/6=1/2+1/3$ then there are no other $a,b$ such that $5/6=a/2+b/3$ – Miles Johnson Nov 21 '19 at 04:44

1 Answers1

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No, this is not true. We can consider the simple case where $t=pq$ where $p,q$ are distinct primes, and $s=1$. Here, we choose $s_p, s_q$ to satisfy $ps_q + qs_p=1$, and such $s_p, s_q$ exist per Bezout's identity. Then observe that

$$\frac{s_p}{p} + \frac{s_q}{q} = \frac{qs_p + ps_q}{pq} = \frac{1}{pq} = \frac{s}{t}$$

In general, numerous $s_q, s_p$ can be found which satisfy $ps_q + qs_p=1$. For instance, take $s_p'=s_p-pk$, $s_q' = s_q+qk$ where $k$ is an arbitrary integer, so that $ps_q' + qs_p' = p(s_q+qk) + q(s_p-pk)=ps_q + qs_p = 1$.

MathematicsStudent1122
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  • If we require that $|s_p|<p$, this should be unique, right? Also, does Bezout's identity imply existence in all cases? – Miles Johnson Nov 21 '19 at 05:36
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    @MilesJohnson If $s=1$ and $t=\prod p^{i_p}$ is arbitrary, in that case it should work out if you add the additional constraint that $|s_p| < p^{i_p}$. Define $a_p = \frac{t}{p^{i_p}}$ and note $\gcd {a_p}_{p|t} = 1$. We can use the extended Euclidean algorithm (generalized to more than two variables) to extract the minimal Bezout coefficients, and they will be unique. To answer your question, yes, Bezout's identity implies existence so long as the numbers are coprime, and if $p,q$ are distinct primes this condition is surely met. – MathematicsStudent1122 Nov 21 '19 at 06:04