Solve for r,t in the Polynomial $3x^3+rx^2+sx+t=0$ with a,b,c Prime

[Albert1]

$a,b,c \in N$
$c+1=2a^2$
$c^2+1=2b^2$
c is a prime
a,b,c are roots of $ 3x^3+rx^2+sx+t=0 $
please find r and t

 

[mathworker]

Re: find f and t

It may not be an ellegent one i just gave it a shot$$
\begin{align*}c^2+2c+1&=4a^4\\c&=2a^4-b^2\end{align*}$$
I gave a try at a=2 so b is 5 and c is 7.
And 7 is also the first prime to satisfy both equations even individually too.
So r is -52 and t is -210

 

[Opalg]

Re: find f and t

$a,b,c \in N$
$c+1=2a^2$
$c^2+1=2b^2$
c is a prime
a,b,c are roots of $ 3x^3+rx^2+sx+t=0 $
please find r and t

[sp]One solution is $a=2$, $b=5$ and $c=7$. Since $a+b+c = -r/3$ and $abc = -t/3$, that gives $r=-42$ and $t=-210$. But is that the only solution?

I think that it is, but I can't prove it. In fact, the equation $c^2+1=2b^2$ can be written $\bigl(\frac cb\bigr)^2 + \bigl(\frac 1b\bigr)^2 = 2$, so that $c/b$ is a lower convergent for $\sqrt2$. The condition $c+1=2a^2$ requires that $\frac12(c+1)$ is a square, and the only Pell number less than $10^9$ satisfying that condition is $7$. So any other solution for $c$ must be bigger than that.[/sp]

 

[Albert1]

Re: find f and t

$a,b,c\in N,\,\, c:a\,\, prime$
$c+1=2a^2---(1)$
$c^2+1=2b^2---(2)$
(2)-(1):$c(c-1)=2(b-a)(b+a)---(3)$
from (1)(2)(3) we conclude :
$c\geq b \geq a,\,\, and \,\,\,\, c \mid (a+b),( or\,\, (a+b) \,\, mod \,\, c=0)--(4)$
$\therefore c=a+b, \rightarrow b=c-a$
from (3)
$c-1=2b-2a=2(c-a)-2a=2c-4a$
$\therefore c+1=4a=2a^2$
$\therefore a=2,c=7,b=5$
$\therefore r=-3\times (2+5+7)=-42,\,\, t=-3\times(2\times5\times7)=-210$

 

[CellCoree]

To solve for r and t, we can use the fact that a,b,c are roots of the polynomial, meaning that when we substitute them into the equation, the equation will equal 0.

Substituting a into the equation, we get:
$3a^3+ra^2+sa+t=0$

Substituting b into the equation, we get:
$3b^3+rb^2+sb+t=0$

Substituting c into the equation, we get:
$3c^3+rc^2+sc+t=0$

We can use this information to create a system of equations:
$3a^3+ra^2+sa+t=0$
$3b^3+rb^2+sb+t=0$
$3c^3+rc^2+sc+t=0$

We can then solve for r and t by using any method of solving systems of equations, such as substitution or elimination.

Using substitution, we can solve for r:
From the first equation, we can solve for t:
$t=-3a^3-ra^2-sa$
Substituting this into the second equation:
$3b^3+rb^2+sb-3a^3-ra^2-sa=0$
Simplifying:
$3b^3+rb^2+sb-3a^3-ra^2=0$
Factoring out r:
$r(b^2-a^2)+s(b-a)=0$
Since a and b are distinct primes, we know that b-a cannot equal 0. Therefore, we can divide both sides by b-a:
$r=\frac{-s}{b-a}$

Now, using this value of r, we can solve for t by substituting it into the first equation:
$t=-3a^3-\frac{s}{b-a}a^2-sa$
Simplifying:
$t=-3a^3-\frac{s}{b-a}a^2-sa$
$t=-3a^3-\frac{s}{b-a}a^2-\frac{sa(b-a)}{b-a}$
$t=-3a^3-\frac{s}{b-a}a^2-\frac{sab-sa^2}{b-a}$
$t=-3a^3-\frac{sab}{b-a}$
$t=-\frac{3a^3(b-a)+s