Find a polynomial f(x) with all coefficients in {0, 1, ..., 9} such that f(2) is prime but f(x) is reducible i.e., f(x) = g(x)h(x) with g, h in of integer coefficients.
@Nick Why so joy?
I am dying trying to approach it.
I can make f(1) prime, that's a simple matter. What about f(2)?
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Visualize this : you have regions $D_1$ and $D_2$, overlapping $D$, not necessarily the whole $D_1 \cap D_2$, but $D \cap D_1 \cap D_2$ is nonnull.(say)
@BalarkaSen Someone must have tutored you at that stage. You mean you studied cubic equations when your mates were learning adding numbers. That is not ordinary, really un ordinary.
We will use the combinatorial identity, which can be proved through induction
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n...
Given that $k^2+k+n$ is always prime for all positive integer $k$ in the interval $\left(0,(n/3)^{1/2}\right)$. Find the largest interval for which the same can be stated.
Do not use HST
(that does not mean Honolulu standard time, okay?)
We will use the combinatorial identity, which can be proved through induction
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n...
