Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?
Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.
Answer is $n(n+1)+an+b$
On
I have searched a way to simplify it but it seems we have to go brute force anyway.
Examination of highest power terms suggests $m=n(n+1)+...$
Then the difference $f(n)f(n+1)-f(n(n+1))=2an^3+...$
Suggesting again $m=n(n+1)+an+...$
Finally you get to factor $b$ in $f(n(n+1)+an)-f(n)f(n+1)$ to get the final result.
An alternative method could be this one, but it seems a bit artificial (knowing the result and all...)
The good thing about it, is that is doesn't require tedious expansions.
$\begin{align}f(n)f(n+1) &=f(n)^2+f(n)f(n+1)-f(n)^2\\\\ &=f(n)^2+\bigg[f(n+1)-f(n)\bigg]f(n)\\\\ &=f(n)^2+(2n+a+1)f(n)\\\\ &=f(n)^2+2nf(n)+af(n)+\overbrace{f(n)}^{n^2+an+b}\\\\ &=\bigg[f(n)^2+2nf(n)+n^2\bigg]+a\bigg[f(n)+n\bigg]+b\\\\ &=f(f(n)+n)\end{align}$
Thus $m=f(n)+n$
You can write this quadratic function in factor form:
$$f(x)= (x-x_1)(x-x_2)$$ where $x_1,x_2$ are roots of $x^2+ax+b=0$ so $$x_1+x_2 =-a\;\;\;{\rm and}\;\;\;x_1x_2 =b$$
Now $$f(n)f(n+1) = (n-x_1)(n-x_2)(n+1-x_1)(n+1-x_2) =\color{red}{(n-x_1)(n+1-x_2)}\color{blue}{(n+1-x_1)(n-x_2)}$$
Since $$\color{red}{(n-x_1)(n+1-x_2)} = n(n+1)-n(x_1+x_2)+x_1x_2-x_1$$ $$=\underbrace{n(n+1)+na+b}_{=:m}-x_1$$ and $$\color{blue}{(n-x_2)(n+1-x_1)} = n(n+1)-n(x_1+x_2)+x_1x_2-x_2$$ $$=\underbrace{n(n+1)+na+b}_{=m}-x_2$$ So if $m=n(n+1)+na+b$ we have $$f(n)f(n+1)=f(m)$$