Problem 7 of Third round

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Problem 7 of Third round

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Source: IV International Festival of Young Mathematicians Sozopol 2013, Theme for 10-12 grade

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Pinko

#1 h Jan 24, 2020, 7:38 AM • 1 Y

Y by Adventure10

Let $a,b,c,$ and $d$ be real numbers and $k\geq l\geq m$ and $p\geq q\geq r$ . Prove that
$f(x)=a(x+1)^k (x+2)^p+b(x+1)^l (x+2)^q+c(x+1)^m (x+2)^r-d=0$
has no more than 14 positive roots.

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#2 h Jan 25, 2020, 6:00 AM • 2 Y

Y by Pinko, Adventure10

Suppose to the contrary. Here we consider the zero polynomial as a polynomial with degree $0$ .
We get $f'(x)=A(x)(x+1)^{k-1}(x+2)^{p-1} +B(x)(x+1)^{\ell-1}(x+2)^{q-1}+C(x)(x+1)^{m-1}(x+2)^{r-1}$ for linear polynomials $A,B,C$ has at least $14$ positive roots.
So does $g(x)=A(x)(x+1)^{k-m}(x+2)^{p-r}+B(x)(x+1)^{\ell-m}(x+2)^{q-r}+C(x)$ .
We get $g'(x)=P(x)(x+1)^{k-m-1}(x+2)^{p-r-1}+Q(x)(x+1)^{\ell-m-1}(x+2)^{q-r-1}+C'(x)$ for quadratic polynomials $P,Q$ has at least $13$ positive roots. (It could happen that $P$ is quadratic when $k=m$ or $p=r$ . Similar special cases determine the degree of other polynomials defined from $Q$ on.)
We get $g''(x)=R(x)(x+1)^{k-m-2}(x+2)^{p-r-2}+S(x)(x+1)^{\ell-m-2}(x+2)^{q-r-2}$ for cubic polynomials $R,S$ has at least $12$ positive roots.
So does $h(x)=R(x)(x+1)^{k-\ell}(x+2)^{p-q}+S(x)$ .
We get $h'(x)=T(x)(x+1)^{k-\ell-1}(x+2)^{p-q-1}+S'(x)$ for a quartic polynomial $T$ has at least $11$ positive roots.
Taking derivatives three more times gives $h''''(x)=U(x)(x+1)^{k-\ell-4}(x+2)^{p-q-4}+0$ for a polynomial $U$ of degree at most $7$ has at least $8$ positive roots, which is impossible.

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#3 h Jan 8, 2022, 11:25 PM

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This problem is false for stupid reasons in some edge cases like $a=b=c=d=0$ or $k=l=m=p=q=r=0$ . Can we impose additional conditions for which @ThE-dArK-lOrD's argument becomes fully correct?

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