If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n.

Question

If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.

I did and I got $$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?

Answer 1

Note that $m$ and $n$ have to be non-negative integers, and perhaps it is implied in the question that $m$ and $n$ are positive as well. The only positive solution to $3n+2m=12$ is $n=2,m=3$, so $m+n=5$.

If zero powers are allowed, then we also have $n=4,m=0$ and $n=0,m=6$, for sums of $4$ and $6$ respectively.

Answer 2

In the first factor the multiplicity of the solution $x=2\vee x=-2$ is $3$, so here we have $6$ solutions. In the second case the solution $x=-1$ has $n$ multiplicity, while in third factor there are solutions: $x=-2 \vee x=-3$ with multiplicity $m$. We have to have: $$3n+2m=12$$ So, every $(m,n)\in N$ such that $3n+2m=12$ are correct. In particular this is a diphantine equation with solutions: $$m=6 \land n=0 \lor m=3 \land n=2 \lor m=0 \land n=4$$

Answer 3

$2×3+3×n+2×m = 18$

Find $m+n$

$3*n+2*m = 12$ ( Diophantine equation )

Offcourse $m,n € Z$ were $Z → (0,1,2,3,4,5,............)$

If we construct a table,

Put $n=0$, $m=6$

Put $n=1$, $m$ ≠ $Z$

Put $n=2$, $m=3$

Put $n=3$, $m$ ≠ $Z$

Put $n=4$, $m=0$

So there are only there values that works here $n=0$, $m=6$ and $n=2$, $m=3$ and $n=4$, $m=0$

Therefore $m+n$ is either $6$ or $5$ or $4$