Divisibility #1

Let $f(x)=x^2+ax+b$ , according to the first condition.

Then $f(-1)=1-a+b>1$ , which means $a<b$ , and let's figure out that $f(1)=1+a+b<1+2b$ for later use.

Also, $f(0)=b$ is a prime number.

If $f(n)$ is a multiple of $k$ for every natural number $n$ , then we can substitute $n=k$ , $n=k-1$ and $n=1$ , and then find out some things.

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(1) $n=k$

$\begin{aligned} &k | f(k)=k^2+ak+b=k(k+a)+b \\ \\ & \therefore \quad k | b \end{aligned}$

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(2) $n=k-1$

$\begin{aligned} &k | f(k-1)=(k-1)^2+a(k-1)+b=k(k-2+a)+1-a+b \\ \\ &\therefore\quad k | 1-a+b \end{aligned}$

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(3) $n=1$

$k | f(1)=1+a+b$

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From (2) and (3), we can figure out that

$k | 2+2b$

And substituting (1), we can say that

$k | 2$

Since $k>1$ , $\boxed{k=2}$ .

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Then, from (1), we know that $b$ is an even number.

We've found that $f(1)<1+2b$ , and we know that $b$ is a prime.

Therefore, $\boxed{b=2}$ .

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What about $a$ ?

Since $2=k|1+a+b=3+a$ and $a<b=2$ ,

$a$ is an odd number which is less than $2$ .

$f(1)+k=1+a+b+k=5+a$ .

So in order for $f(1)+k$ to have the largest value possible, $\boxed{a=1}$ .

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Therefore, the maximum of $f(1)+k$ is

$f(1)+k=1+a+b+k=\boxed{6}$ .