Show that： for any $k\ge 100,(k\in N^{+})$, there exsit $p\in N^{+}$, such $$\begin{cases} a+b+c=k\\ abc=p\\ a>b>c \end{cases}$$ has at least two postive integers solution $(a,b,c)$
This links some people say use mathematical induction,But I think can't solve it.so I use Vieta's formulas consider $$x^3-kx^2+() x-p=0$$ What approaches do you think I could take to solving the next step？
This problem is from when I solve this following Mathmematician 2014:
Hint/Short answer : Look for pairs of triples of the form $(x,2y,3z),(6x,y,z)$.
Longer answer: I can show that the result already holds for $n\geq 83$ (also, a computer search shows that the largest $n$ for which there is no solution is $n=22$). Indeed, for such $n$ let us put $u=\frac{n+1}{22}$ and $v=\frac{n-1}{17}$. We have $$v-u=\frac{5n-39}{374} \geq \frac{376}{374} > 1,$$ so that there is a (necessarily positive) integer $a$ between $u$ and $v$. Then $\frac{n}{22} < a <\frac{n}{17}$, and we may consider the triples $p=(2a,4(n-17a),3(22a-n))$ and $p'=(2(n-17a),22a-n,12a)$.
All the entries in $p$ are distinct, except when $n$ is divisible by $35,64$ or $134$.
All the entries in $p'$ are distinct, except when $n$ is divisible by $23$ or $56$.
The triplets $p$ and $p'$ are distinct, except when $n=18a$ or $n=20a$ (indeed, if we put $\sigma_2(x,y,z)=xy+xz+yz$, we have the identity $\sigma_2(p)-\sigma_2(p')=-10(n-18a)(n-20a)$).
So, if $n$ is divisible by none of $18,20,23,34,35,56,64$ or $134$, the triplets $p$ and $p'$ satisfy all that we want and we are done.
The remaining cases can be done by hand :
If $n$ is divisible by $20$, $n=20t$, consider triplets $q=t(1,9,10)$ and $q'=t(2,3,15)$.
If $n$ is divisible by $23$, $n=23t$, consider triplets $q=t(4,9,10)$ and $q'=t(5,6,12)$.
If $n$ is divisible by $35$, $n=35t$, consider triplets $q=t(1,14,20)$ and $q'=t(2,5,28)$.
If $n$ is divisible by $56$, $n=56t$, consider triplets $q=t(1,23,32)$ and $q'=t(2,8,46)$.
If $n$ is divisible by $64$, $n=64t$, consider triplets $q=t(1,30,33)$ and $q'=t(3,6,55)$.
If $n$ is divisible by $134$, $n=134t$, consider triplets $q=t(1,45,88)$ and $q'=t(3,11,120)$.
The only case left now is the case when $n=18a$ (so that $p$ and $p'$ coincide).
If $a$ is divisible by $2$, $a=2s$, consider triplets $q=s(2,13,21)$ and $q'=s(3,7,26)$.
If $a$ is divisible by $3$, $a=3s$, consider triplets $q=s(1,18,35)$ and $q'=s(2,7,45)$.
If $a$ is divisible by $5$, $a=5s$, consider triplets $q=s(1,24,65)$ and $q'=s(2,10,78)$.
If $a$ is divisible by $7$, $a=7s$, consider triplets $q=s(1,29,96)$ and $q'=s(4,6,116)$.
Otherwise, we must have $a\geq 11$, so $n\geq 18\times 11$ and reasoning as above we see that $v-u\geq 2$ so there are at least two integers between $u$ and $v$. We can then replace $a$ (such that $18a=n$) with an $a'$ such that $n\neq 18a'$, which concludes the proof.