anantmudgal09 wrote: Find all integers $k$ such that all roots of the following polynomial are also integers: $$f(x)=x^3-(k-3)x^2-11x+(4k-8).$$ Let $x$ an integer root to this polynomial. So obviously $|x| \neq 2$ and we have: \[k = \frac{x^3+3x^2-11x-8}{x^2-4}= x+3+\frac{4-7x}{x^2-4}\]which means that $x^2-4$ divides $4-7x$: \[(4-7x)^2-(x^2-4)^2= -x(x+8)(x-1)(x-7) \geq 0\]We conclude that: $-8\leq x \leq 7$ A quick verification leads to $k=5$ the only solution to this problem and in this case \[f(x)=(x-1)(x-4)(x+3)\]

$f(x)=x^3-(k-3)x^2-11x+(4k-8)=(x-a)(x-b)(x-c)$ $f(2)=8-4(k-3)-22+4k-8=-10=(2-a)(2-b)(2-c)$. So we have conditions: $(2-a)(2-b)(2-c)=-10; ab+ac+bc=-11$ Solving in integers we find $(a,b,c)=(1,4,-3)$. $abc=8-4k=-12 \to k=5$ $a+b+c=k-3=-2 \to k=5$ so $k=5$ is solution.

RagvaloD wrote: $a+b+c=k-3=-2 \to k=5$ so $k=5$ is solution. $a+b+c=k-3=2$

Observe that the value of $f(2) $ is negative , while that of $f(-2)$ is positive . As the function given is continuous, by Intermediate Value Theorem , there must exist a root of $f(x) $ between $-2$ and $+2$ . As we are given the roots are integers , we just check the roots $ -1, 0, 1$ ... We see that we get $k=5 $ from these values as the only solution for which the given equation has integral roots ...

AdBondEvent wrote: Observe that the value of $f(2) $ is negative , while that of $f(-2)$ is positive . As the function given is continuous, by Intermediate Value Theorem , there must exist a root of $f(x) $ between $-2$ and $+2$ . As we are given the roots are integers , we just check the roots $ -1, 0, 1$ ... We see that we get $k=5 $ from these values as the only solution for which the given equation has integral roots ... Nice!

AdBondEvent wrote: Observe that the value of $f(2) $ is negative , while that of $f(-2)$ is positive . As the function given is continuous, by Intermediate Value Theorem , there must exist a root of $f(x) $ between $-2$ and $+2$ . As we are given the roots are integers , we just check the roots $ -1, 0, 1$ ... We see that we get $k=5 $ from these values as the only solution for which the given equation has integral roots ... Quote: Too good

bel.jad5 wrote: $-8\leq x \leq 7$ $ x \in [-8, 0] \cup [1,7] $

Apply viete and $a^2+b^2+c^2\ge (ab+bc+ac)$ Interestingly this q also In previous year iit jee

Let the roots of the cubic polynomial be $a,b,c$. Note that, $\sum ab$ is odd and $abc$ is even. This is possible only when exactly one of $a,b,c$ is even and the other two must be odd. WLOG, let $a$ is even and $b,c$ are odd. Thus, $a+b+c\equiv k-3\equiv 0\pmod{2}\implies k$ is odd. Thus, $abc=4k-8=4\times \text{odd}\implies a=\pm 4$. Substituting $x=\pm 4$ in $f(x)=0$, we get $k=5$ is the only possible value for $k$ which indeed results $$f(x)=(x-1)(x-4)(x+3)$$

The above solution is wrong. $a$ is even doesn't imply that $a=\pm 4$. $a$ could have been something like $4p$ where $p\mid bc$.

AdBondEvent wrote: Observe that the value of $f(2) $ is negative , while that of $f(-2)$ is positive . As the function given is continuous, by Intermediate Value Theorem , there must exist a root of $f(x) $ between $-2$ and $+2$ . As we are given the roots are integers , we just check the roots $ -1, 0, 1$ ... We see that we get $k=5 $ from these values as the only solution for which the given equation has integral roots ... Your IVT solution is very nice. My noob solution for storage:

With $a,b,c$ being our roots, notice that $$a+b+c=k-3$$$$ab+bc+ac=-11$$$$abc=8-4k$$ So, $(a+2)(b+2)(c+2)= abc+2(ab+bc+ac)+4(a+b+c)+8=-18.$ Now the rest is factor and case work (observations about parity may also help cut this).