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#1 h Mar 22, 2011, 1:15 PM • 1 Y

Y by Adventure10

A monic quadratic polynomial $f$ with integer coefficients attains prime values at three consecutive integer points.show that it attains a prime value at some other integer point as well.

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tenniskidperson3

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tenniskidperson3

#2 h Mar 23, 2011, 12:19 AM • 2 Y

Y by Adventure10, Mango247

We know that if $f(x)=x^2-bx+c$ then $f(a)=f(b-a)$ . Thus if the consecutive integers $x-1, x, x+1$ are not $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$ then one of the three consecutive numbers is at least a distance of $\frac{3}{2}$ from $\frac{b}{2}$ . Let this number be $y$ . Then

$f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$ .

Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. So we have another integer input to the function with a prime output.

So now we assume that the 3 numbers are in fact $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$ . These are integers only if $b=2k$ or $f(x)=x^2-2kx+c=(x-k)^2+(c-k^2)$ . We know that at $k-1, k, k+1$ these are primes. But

$f(k+1)-f(k)=((k+1)-k)^2+(c-k^2)-(k-k)^2-(c-k^2)=1$ .

The only two primes that are 1 apart are $2$ and $3$ . Thus $f(k)=2$ , so $c-k^2=2$ and $f(x)=(x-k)^2+2$ . Then $f(k+3)=(k+3-k)^2+2=11$ which is another prime value at an integer point.

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#3 h Mar 23, 2011, 2:03 AM • 1 Y

Y by Adventure10

tenniskidperson3 wrote:

... Let this number be $y$ . Then

$f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$ .

Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. ...

Why must $b-y$ be one of the 3 consecutive integers?

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tenniskidperson3

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#4 h Mar 23, 2011, 6:15 PM • 2 Y

Y by Adventure10, Mango247

There is no reason why it should be; in fact it isn't. I don't understand what you're saying; $b-y$ is at least 3 away from $y$ which is one of the consecutive numbers, so $b-y$ cannot be.

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#5 h Mar 24, 2011, 1:09 AM • 2 Y

Y by Adventure10, Mango247

Suppose $f(n-1),f(n),f(n+1)$ are primes. Define another polynomial $g(x)=f(x+n)$ . So $g$ is a monic quadratic polynomial with integer coefficients such that $g(-1),g(0),g(1)$ are primes. It is enough to show that $g(x)$ attains a prime value at another integer point. Let $g(x)=x^2+ax+b$ . Note that $g(x)=g(-a-x)$ . So $g(-a-1),g(-a),g(-a+1)$ are primes. If the set $\{-a-1,-a,-a+1\}$ does not coincide with $\{-1,0,1\}$ , then we're done. Otherwise, we get $a=0$ . Then $g(-1)=1+b$ and $g(0)=b$ , both of which are primes. So $b=2$ and $g(x)=x^2+2$ , which attains a prime value at $g(3)=11$ .

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