Ok I think I got it. There are two important facts:

1-the q are always positive so there is a restriction on the possible distributions they might assume

2-their sum is a chi square with r degrees of freedom, and the chi square is always positive and assumes every possible value from 0 inlcuded to +infinity.

So if we put it this way:

[B(s)]^r2=[Chisq(s)/A(s)]^r1*[Chisq(s)]^r2

Let's call Chisq(s)/A(s)=D(s)

We know that this product is the mgf of a convolution of a chisquare with r2 dfs and an unknown distribution d(x), whose final result must be non negative.

Now, since the second term of the convolution assumes every possible value from 0 to + infinity, then d(x) must be defined only for nonnegative values, otherwise their convolution (sum of the two random variables) could take negative values.

So D(s) is the mgf of the distribution of a non negative value.

Now let's examine D(s)=Chisq(s)/A(s). It means it is the mgf of the distribution of the difference between a random variable distributed like a chisquare with 1 df and another random variable with unknown distribution a(x).

Now, even if a(x) is defined only for nonnegative values, the difference between it and the chisquare could take negative values (since the chisquare could assume any value between 0 and + infinity), UNLESS they coincide exactly, which means their difference is always 0, the distribution is a dirac delta centered in 0 and the mgf is therefore 1.

So A(s)=Chisq(s), and B(s)=Chisq(s), Q.E.D.

Thanks again for the hint!

Btw this proof is more general than one with linear algebra, because we can also assume non-integer dfs!