Can this product of sigma sums be reduced further What I have so far:
$$3\sum_{k=0}^{2j} x^{k}\cdot\sum_{k=0}^{2j_1} (x+\lambda_1)^{k} \dots\cdot\sum_{k=0}^{2j_m} (x+\lambda_m)^{k}\quad{-} \quad 4(x^{2j})(x+\lambda_1)^{2j_1}\dots(x+\lambda_m)^{2j_m}+1=0 $$
Why I'm stuck: I don't know if the LHS of the minus can be simplified and I know that the RHS of the minus can be simplified using the multi binomial theorem but I have a fixed x. Any help is appreciated.
The multi binomial theorem says that:
$$(x_1+\lambda_1)^{n_1}\dots(x_m+\lambda_m)^{n_m}=\sum_{k_1=0}^{n_1}\cdot \cdot\cdot \sum_{k=0}^{n_m} {n_1 \choose k_1 }x_1^{k_1}\lambda_1^{n_1-k_1} \cdot \cdot \cdot {n_m \choose k_m}x_m^{k_m}\lambda_m^{n_m-k_m}$$ so I am just kinda trying to first try something on the RHS of the minus sign first.
 A: If the question is really asking about the equation $3\sum\ldots - 4\prod\ldots=0$, it's easy to see there are no solutions in integers:
$$\sum_{k=0}^{2m} z^k = 1 + z(z+1)\sum_{i=0}^{m-1} z^{2i}$$
Since $z(z+1)$ is always even, each sum is odd and so is their product. Clearly, $3$ times an odd number cannot be equal to an integer multiplied by $4$.
A: Attempt:
Maybe: $$3\sum_{k_0=0}^{2j} x^{k}\cdot\sum_{k_1=0}^{2j_1} (x+\lambda_1)^{k_1} \dots\cdot\sum_{k_m=0}^{2j_m} (x+\lambda_m)^{m} \quad - \quad 4\sum_{k_1=0}^{n_1}\cdot \cdot \cdot \sum_{k_m=0}^{n_m} {n_1 \choose k_1 }x_1^{k_1}\lambda_1^{n_1-k_1} \cdot \cdot \cdot {n_m \choose k_m}x_1^{k_m}\lambda_m^{n_m-k_m}$$ where all I have done is rewritten RHS   of the minus sign using multi binomial theorem. Then I can attack the LHS by treating the terms as binomials so : $$3\sum_{k_0=0}^{2j}x^k\cdot \left( \sum_{k_1=0}^{2j_1}\sum_{r=0}^{k_1} {k_1 \choose r}x^r \lambda_1^{k_1-r} \right) \dots \left( \sum_{k_m=0}^{2j_m}\sum_{r=0}^{k_m} {k_m \choose r}x^r \lambda_1^{k_m-r} \right) \quad - \quad 4\sum_{k_1=0}^{n_1}\cdot \cdot \cdot \sum_{k_m=0}^{n_m} {n_1 \choose k_1 }x_1^{k_1}\lambda_1^{n_1-k_1} \cdot \cdot \cdot {n_m \choose k_m}x_1^{k_m}\lambda_m^{n_m-k_m}$$
and then rearrange the RHS a such: $$3\sum_{k_0=0}^{2j}x^k\cdot \left( \sum_{k_1=0}^{2j_1}\sum_{r=0}^{k_1} {k_1 \choose r}x^r \lambda_1^{k_1-r} \right) \dots \left( \sum_{k_m=0}^{2j_m}\sum_{r=0}^{k_m} {k_m \choose r}x^r \lambda_1^{k_m-r} \right) \quad - \quad 4\sum_{k_1=0}^{n_1}\left( {n_1 \choose k_1} x_1^{k_1}\lambda_1^{n_1-k_1} \right)\cdot \cdot \cdot \sum_{k_m=0}^{n_m}  {n_m \choose k_m}x_1^{k_m}\lambda_m^{n_m-k_m}$$
