1

Is that true that every number greater than a product of relatively prime numbers is a linear combination of those prime numbers? Why would that be a case? A teacher told us in class today that such a thing works, but I was not able to determine if it's true and if so, how is it called.

thefool
  • 1,024
  • 2
    Cf. $n=2$ case of Frobenius Coin Problem. – John Omielan Nov 03 '22 at 05:05
  • 1
    Can a linear combination involve subtraction? If so every integer is a linear combination (Bezout's lemma/Euclid's algorithm shows you that the greatest common divisor is a linear combination and so if they are realtively $1$ is and thus every integer is). The smaller combos will require that some terms must be subtracted. If the terms must be addition it must be large enough that you don't have to subtract. Frobenius Coin goes into detail but basically if its large enough to be the product you can add more of one and subtract more the other without dipping into negatives. – fleablood Nov 03 '22 at 05:34
  • If your numbers are $p,q$ then Bezout's Lemma tells us we can find $ap -bq=1$ for some $a,b$. But this uses negative value. But If the number we want is $M=pq + k>pq$ for a non negative $k$ we can use Bezout to get $cp -dq =k$ so $pq+cp-dq=cp+(p-d)q=pq+k$. We have to futz a bit but we can show that we can get $cp-dq=k$ with $d< p$ so $cp+(p-d)q= M$ with $c,p-d> 0$ and we can avoid negatives. Try an example and it may come clear. – fleablood Nov 03 '22 at 05:44

0 Answers0