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The following problem has popped up during my master's project, I am unsure how to solve it completely and I feel like there must be a nice argument to do it:

Given polynomials:

  1. $p_{a,b}(x,y) = -2abx^{a-1}y^{b-1} + x^{a+1}y^b $
  2. $q_{c,d}(x,y) =cx^{c-1}y^d$
  3. $r_{f,g}(x,y) = -x^fy^g$

find all $\mathbb{R}-$linear combinations of them which give $0$. The $a,b,c,d,f,g \in \mathbb{Z}_{\geq 0}$ can be arbitrarily chosen. We let $ \lambda x^{\lambda -1} = 0$ when $\lambda = 0$.

I have found lots of combinations by brute force, for example, letting $f=c-1$ and $g=d$ we see that $q(x,y) + c \cdot r(x,y) = 0$, but I am struggling to classify all the combinations to ensure I have them all.

Edit: Note that you're able to take one of the given polynomials multiple times and use different a,b values, so for example I could start with $$p_{a,b}+\frac{1}{2(a+2)(b+1)}p_{a+2,b+1}$$ to give $$−2abx^{a−1}y^{b−1}+\frac{1}{2(a+2)(b+1)}x^{a+3}y^{b+1}$$ (assuming my calculations are correct) and then use some combinations of $q$ and $r$ to kill this off.

Any suggestions on what areas of maths and/or strategies could help me solve this would be much appreciated.

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  • $\begingroup$ Representation of polynomial is unique, i.e. if a linear combination is zero then all "monomials" have to be zero. In your case if $A*p_{a,b} + Bq_{c,d} + Cr_{f,g} = 0$ then you just consider different cases like, for example $A2ab = Bc$ when $a-1 = c-1$ and $b-1 =d$. etc. $\endgroup$
    – Salcio
    Feb 18 at 0:07
  • $\begingroup$ @Salcio Sorry I haven't worded my question very well, you are able to take one of the given polynomials multiple times and use different $a,b$ values, so for example I could take $p_{a,b} + \frac{1}{2(a+2)(b+1)}p_{a+2,b+1}$ to give $-2abx^{a-1}y^{b-1} + \frac{1}{2(a+2)(b+1)}x^{a+3}y^{b+1}$ (if my calculations are correct) and then use $q$ and $r$ to kill this off. I will edit this in, thanks $\endgroup$
    – Dylan
    Feb 18 at 0:43

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