For starters, there is a trivial answer. Let c=lcm(a,b), then c=xa=yb. The last two subexpressions xa=xa+0b and yb=0a+yb are trivial linear combinations of c.

More generally: let c=lcm(a,b), d=gcd(a,b), then c = ab/d = a*k*(b/d)-b*(k-1)*(a/d), where k is an arbitrary integer.

EDIT: updated c=ab/d ("/d" was missing), and added the constraint on k to be an integer (thanks for the remarks, Madline Ali Al- Tahan and Raul Simon)

Since a divides lcm(a,b), lcm(a,b)=ma +0b. Similarly, lcm(a,b)=0a+nb. Further, lcm(a,b)=a .(b-x) (putting m=b-x) so that lcm(a,b)=ab-xa another linear combination and so on (infinitely many ways).   Conversely, if ma+nb=lcm(a,b), then b divides m.gcd(a,b) and a divides n.gcd(a,b).

Madline, I am sorry, but it seems that my previous answer (I had deleted it but I wrote it here again now to illustrate this comment) was wrong because I did not realize the coeficients c1/2 and c2/2 of the linear combination on a and b had to be integers in your problem. So it only works sometimes. It is not so easy...Sorry and good luck :-)

Previous answer:

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I think it is possible because for some integer constants c1 and c2, where c1=lcm(a,b)/a  and c2=lcm(a,b)/b we have

lcm(a,b)= c1*a     (1)

and also

lcm(a,b)=c2*b      (2)

So we add (1) and (2) and get

2*lcm(a,b) = c1*a + c2*b

and finally

gcd(a,b)=(c1/2)*a + (c2/2)*b

which is a linear combination on the integers a and b.

Please check it.

Good luck.

Is it possible to write lcm(a,b) as linear combination of a and b?. Available from: https://www.researchgate.net/post/Is_it_possible_to_write_lcma_b_as_linear_combination_of_a_and_b#view=55363d73f079edf07e8b45d4 [accessed Apr 21, 2015].

"

   Emil, I surmise that your answer is based on the fact that cd = ab, and k = arbitrary integer > 1. (Nice solution.)

In general. Let R be an integral domain, gcd(a, b)  and lcm(a,b) be  linear combination of a and b in R, [see Bezout's Identity and see Ritumoni and Emil advice above respectively]  if and only if R is a Bezout domain, i.e., every finitely generated ideal of R is principal.

If you are merely asking for linear combinations, it is easy to do it, as noted above.

However, if you are asking for strictly positive integer linear combinations, things are much less simple: we can find a very simple example (a=2, b=3) in which there is no strictly positive linear combination of them giving lcm(a,b)=6.  

You might look at references to the coin problem (WP has a good write up here:

http://en.wikipedia.org/wiki/Coin_problem ): much is known in the case of two variables, much less in the case of many variables.  Note that this problem is about giving non-negative solutions, which corresponds in my example to giving non-negative solutions to ax+by=ab-a-b (=1).

Neil

Ok thank you Prof. Neil. This is exactly what I want.

Are there any publication on lcm(a, b)=ax+by where a and b are non zero integers?

Emil: c is not equal to a*b unless a and b are relatively prime.

Thank you Madline. I've corrected it. It is c=ab/d and the rest should be correct.

Madline,

There is a wealth of literature on when you can solve ax+by=n in positive integers.  It is clearly enough (in the two variable, a, b, case) to consider the case when they are coprime.  Then it is known that the largest integer not expressible as a non-negative integer linear combination is ab-a-b.  Hence the largest integer not expressible as a positive integer linear combination is ab.  The wiki page I linked to above deals with this case under the section "n=2".  For more variables than 2, things get much more complicated.

Neil

Dear Madline,

Please see our proof in the attachment.

The aim of this question is to check whether there are researches on writing lcm of a and b as positive linear combination of a and b. And if you read the answers written by experts, you may conclude that it is not that easy nor trivial task.

Peter --- the Euclidean algorithm will never give positive solutions for gcd(a,b) =ax+by, (when a, b are positive integers) since gcd(a,b)