I hope I have solved it.

Please vet the below proof.

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I want to prove that if there are possibly divisors smaller than g.c.d., they are not linear combinations of the integers involved.

If c is a common divisor, then a = cx, b = cy, ∃x,y∈Z. Then, c is a divisor of the linear combination cx+cy. Here is the fallacy, as how can I mathematically say that a gcd is a linear combination, but not c.

My logic goes as follows:

As earlier stated, a=cx, b=cy. So, need find new multipliers, say ∃ e,f∈Z to prove that c = e.c.x+f.c.y is a linear combination.

This

equation can be reduced to, for c≠0:

1 = ex+fy

So, the options are :

(i) ex=1,fy=0 => both e=±1e=±1, and either f=0, or y=0

(ii) ex=0,fy=1 => both f=±1, & y=±1, and either e=0, or x=0

But, x,y≠1, as these are the multipliers needed to equate c to a,b respectively.

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