How find all values of parameter $p \in R$ such that equation $~ 5x^{3}-5(p+1)x^{2}+(71p-1)x-(66p-1)=0$ has three solutions which are natural numbers?
My try $(x - 1)(5x^2 - 5px + 66p - 1)=0$ and what next?
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Better to put your attempt in the question body. This is a pre-emptive strike dispelling the thoughts (that will otherwise develop) that you are just outsourcing a homework problem. I took the liberty of moving it there. – Jyrki Lahtonen Jul 17 '15 at 14:10
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$5x^3-5(p+1)x^2+(71p-1)x-(66p-1)=(x-1)(5x^2-5px+66p-1)$, so now we need to know when $5x^2-5px+(66p-1)$ has two integer roots.
Let theese roots be $a$ and $b$. Then $a+b=p$, $ab=(66p-1)/5$. So, $5ab=66(a+b)-1$, and we can write this as $25ab-330(a+b)+5=0$, or $(5a-66)(5b-66)=66^2-5=4351=19\cdot 229$.
So, the only opportinities (if we don't look at order of $a$ and $b$) we can have are:
$5a-66=4351$, $5b-66=1$, what doesn't hold for $a,b\in\mathbb{Z}$
$5a-66=19$, $5b-66=229$, which means $p=a+b=17+59=76$.
$5a-66=-19$, $5b-66=-229$, doesn't hold for $a,b\in\mathbb{Z}$.
$5a-66=-4351$, $5b-66=-1$, which means $p=a+b=-857+13=-844$ but in this case $a=-857$ is not a natural number.
So, answer is $p=76$ (if I didn't miscount).
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Note that $x=1$ is always an integer root, so we have $$ 5x^3-5(p+1)x^2+(71p-1)x-(66p-1)=(x-1)(5x^2- 5px + 66p - 1). $$ Hence we can reduce the question to the quadratc polynomial equation $$ 5x^2- 5px + 66p - 1=0. $$ For which real parameters $p$ does it have two integral roots ? Using Vieta we see that there are exactly two possibilities: $p=-844$, with $x=1,13,-857$ as integral roots, and $p=76$, with $x=1,17,59$ as integral roots.
Dietrich Burde
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$\Delta>0$,$25p^2 - 1320p + 20>0$,$\frac{66p-1}{5} \in N$
$p>52$,$p=76$ is only solution?
– piteer Jul 17 '15 at 14:01