Polynomial function

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juandiaz

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a,b are integers.
P(x) = x^2+ax+b,
P(2020 + P(2020) ) = 1.
Find sum of possible a.F00C7BD2-CF73-4BC6-9914-47F399AADA66.jpeg
 
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Please read the posting guidelines.
 
An interesting problem, but I wonder what "sum of possible a" mean? Perhaps the sum of all possible a?

:)
 
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I am quite interested in this problem.
Here is what i have done so far.
I supposed p(r) =1 and solved for r. Then I set r=2020 + p(2020).
I tried other things but nothing need is coming out.
Any hints?
 
Steven G said:
I am quite interested in this problem.
Here is what i have done so far.
I supposed p(r) =1 and solved for r. Then I set r=2020 + p(2020).
I tried other things but nothing need is coming out.
Any hints?
Click to expand...
I did find two solutions but my method involved brute force and intense algebra. Since the question has 2020, it's probably a competition problem. It's likely to have a prettier solution than mine. :(
 
Competition and simple solution--yes
I am sure that I can get the solution but I don't want to use brute force.
Someone, including possibly one of us, will come along with an elegent solution.
Very interesting problem.
 
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AvgStudent said:
I'll share it with you through a private message. I don't want to hijack the thread.
Click to expand...
I feel that this problem is difficult enough where you can post your solution especially since your solution is not the way it needs to be done during a competition. Of course if you are uncomfortable with posting the solution here then please send me a private message.
 
Given [imath]p(x) = x^2+ax+b[/imath], then
\(\displaystyle p(x+p(x))=p(x^2+(a+1)x+b)=(x^2+(a+1)x+b)^2+a(x^2+(a+1)x+b)+b=1\)

After intense algebra, I get...
\(\displaystyle (a x + b + x^2) (a x + a + b + x^2 + 2 x + 1) =1\)

Substituting [imath]x=2020.[/imath]
\(\displaystyle (2020a + b + 2020^2) (2020a + a + b + 2020^2 + 2(2020) + 1) =1\)

Equating the factors of 1 yield 2 systems of equations.
[math]\begin{cases} (2020a + b + 2020^2)=1 \\ (2020a + a + b + 2020^2 + 2(2020) + 1) =1\\ \end{cases}\implies \boxed{a=-4041, b=4082421}[/math]
[math]\begin{cases} (2020a + b + 2020^2)=-1 \\ (2020a + a + b + 2020^2 + 2(2020) + 1) =-1\\ \end{cases}\implies \boxed{a=-4041, b=4082419}[/math]
The sum of all possible [imath]a = -4041[/imath]
 
Does anyone have a better way of solving this problem or maybe some other ideas?
 
Actually that should be p([x) + 2x +a + 1]
I guess that you get a two minute penalty in the penalty box (the corner)
 
Steven G said:
Actually that should be p([x) + 2x +a + 1]
I guess that you get a two minute penalty in the penalty box (the corner)
Click to expand...
To atone for my mistake, it appears that [imath]p(x+p(x))=p(x)p(x+1)[/imath]
So [imath]p(2020 +p(2020))=p(2020)p(2021)=1[/imath] will give you the 2 system of equations @AvgStudent had above.
Not sure why this is true though.
Edit: The identity has been proven on MSE.
math.stackexchange.com

Proving there exists $k$ such that $p(n)p(n+1)=p(k)$

Exact Question Let $p(x)$ be monic quadratic polynomial over $\mathbb{Z}$. Show that for any integer $n$, there exists an integer $k$ such that $p(n)p(n+1)=p(k)$ Expanding the polynomial just c...
math.stackexchange.com math.stackexchange.com
 
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