If $P$ is a monic polynomial of degree $2015$ such that $P(i)=2014-i\quad\forall i=0,…,2014$ & $P(2015)=n!-a$, find $n+a$

If $P(x)$ be a polynomial of degree $2015$ with leading coefficient $1$ such that $P(0)=2014, P(1)=2013, P(2)=2012,...,P(2014)=0$ and $P(2015)=n!-a$, where $n$ and $a$ are natural numbers, find $(n+a)$.

Attempt:

Obviously the constant is $2014$. The sum of the coefficients is $2013$. And it can be observed that $P(x+1)-P(x)=1$

But then what?

• Can it be observed that $P(x+1)-P(x)=1$ for all $x$? If that is true, then we have $P(x)=2014-x$, which contradicts the leading coefficient statement. Try interpolation formulas, those will find the polynomial that matches those points. – Simply Beautiful Art Nov 13 '16 at 14:40
• Hmm. We haven't done interpolation formulae in class yet, so I assume this can be done without that? – Shashank Holla Nov 13 '16 at 14:42
• This question is a tiny bit flawed: if $(n,a)$ is a solution, then so is $(n+k, (n+k)!-n!+a)$ for all $k \in \mathbb N$. So perhaps it should ask for the smallest possible value of $n+a$. – TonyK Nov 13 '16 at 15:10
• @TonyK, It's given that the degree is 2015, so there has to be exactly one solution. If the degree weren't mentioned, we would have a minimum value – Dhanvi Sreenivasan Nov 13 '16 at 15:20
• @Dhanvi: You have missed my point. In this case we can take e.g. $n=2025,a=2025!-2015!+1$, and we still have $P(2015)=n!-a$. – TonyK Nov 13 '16 at 15:55

We define a new function $g(x) = p(x) + x - 2014$.

Now, we know degree of $p(x)$ is 2015. Hence, degree of $g(x)$ is also 2015

Now, $g(x)$ has roots $0,1,2,3...2014$. Hence, we write $g(x)$ as

$$g(x) = x(x-1)(x-2)....(x-2014)$$ $$\implies p(x) +x - 2014 = x(x-1)(x-2)...(x-2014)$$ $$\implies p(x) = x(x-1)(x-2)....(x-2014) -x + 2014$$

Hence, $p(2015) = 2015! - 2015 + 2014 = 2015! -1 \implies n+a = 2016$

• Brilliant! Thanks! – Shashank Holla Nov 13 '16 at 14:43
• This might seem a bit weird, but what made you get that idea? – Shashank Holla Nov 13 '16 at 14:44
• I was a JEE aspirant at one time :P.. I don't really have a concrete reason as to how I got it – Dhanvi Sreenivasan Nov 13 '16 at 14:46
• @Shashank It is obvious what $p(x)$ was for $x\in\{0,1,2,\dots,2014\}$, and since it was a polynomial, he found another polynomial $2014-x$ that matched those points so that his $g(x)=0$ at all $x\in\{0,1,2,\dots,2014\}$, and since the first coefficient was $1$, he factored it simply and solve for $p(x)$. – Simply Beautiful Art Nov 13 '16 at 14:47
• I had exactly the same idea, but @Dhanvi was too quick for me :-( – TonyK Nov 13 '16 at 15:11