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The polynomial $p^3+ap^2+bp+13$ has three integer roots. Find the values of a and b.

I can't come up with anything at all, could someone show me a baby hint please?

Does it mean anything for a cubic to have three integer roots?


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Let the roots be $x,y,z$. Then $p^3 + ap^2 + bp + 13 = (p-x)(p-y)(p-z)$. – Daniel Fischer Sep 8 '13 at 17:37

Just compute $(p-r)(p-s)(p-t)$ and compare coeficients with $p^3+ap^2+bp+13$. This gives Diophantine equations, i.e., $-(r+s+t)=a$, $rs + rt + st=b$, and $rst=-13$. A possible solution is $p^3 + 11p^2 - 25p + 13$, or $p^3 - 13p^2 - p + 13$, etc.

Comment: Yes, it means a lot for a cubic to have three integer roots. It even means a lot to have only real roots.

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Hint: consider the product of the roots. Three roots evidently means three different roots. Given what you know about the product, what could those roots be?

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Finding the roots means finding a value for p that makes the polynomial zero, so one can write:

$$p^3 + a p^2 + b p + 13 = 0 $$

Rewriting a little gives

$$p^3 + a p^2 + b p = -13 \\(p^2 + a p + b)p = -13$$

If it is only integer root you seek, this means $p$ is a factor of $-13$ and $(p^2 + a p + b)$ is its co-factor $c=(-13/p)$, so $c \in \{-13,-1,1,13\}$.

The quadratic $(p^2 + a p + b = c)$ has roots given by $p = \frac{-a \pm \sqrt{a^2 - 4(b-c)}}{2}$, so $$a^2-(2*p+a)^2 = 4(b-c)$$ A guaranteed integer solution occurs when $a=\pm\frac{2(b-c)+2}{2}$ and $2p+a=\pm\frac{2(b-c)-2}{2}$. This occurs in 8 separate cases:

\begin{eqnarray} \pm a &=& b - 12 \\ \pm a &=&b \\ \pm a &=& b+2 \\ \pm a &=& b+14 \end{eqnarray}

Other solutions are possible when $b-c$ is factorable.

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