Question:
[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?
To begin with, your solution is correct, and is actually a quick way to solve this problem. That is, It is efficient.
However, I think what you are trying to say, is there a way to write the solution "more rigorously", and the answer to this is: yes.
Let \(P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0\), where \(a_n,a_{n-1},...,a_0 \ge 0\) (Given.)
Since: \(P(1) = 5 \iff a_n+a_{n-1}+...+a_0=5\), which does not tell us that much.
Next, \(P(P(1)) = P(5) =a_n*5^n+a_{n-1}*5^{n-1}+...+a_0=177\)
Notice: \(5^3=125,5^4=625\); hence, \(P(x)\) is at most a cubic polynomial.
Now,
Let: \(P(x)=ax^3+bx^2+cx+d\) , where \(0\le a,b,c,d \le 5\)
We know: \(a+b+c+d=5\) (1)
And, \(125a+25b+5c+d=177\) (2)
Notice from (2):
\(125a \le 177 \implies a \le 1.416 \implies a=1\text{ or, } a=0\)
Next, consider the case a=1 :
(2) and (1) becomes:
\(25b+5c+d=52 \text{, and } b+c+d=4\)
Using the same idea as before:
\(25b \le 52 \implies b \le 2.09 \implies b=2,1,0\)
But, \(b=1,0\) is rejected. Because observe that if b=0, then 5c+d = 52 (and this is only true for large c and d, which we are not given.)
Proof: \(c,d\le 5 \implies 5c \le 25, d\le 5 \implies 5c+d \le 25+5=30\), so it is at most 30 (Which, by the way, is even unattainable!).
Thus, \(b=2\) only.
Next, we substitute b=2 in (1) and (2) to get:
\(c+d=2\text{ , }5c+d=2 \implies 5c\le 2 \implies c=0,d=2\)
So: \(a=1,b=2,c=0,d=2\) is a valid solution.
Thus, \(P(x)=x^3+2x^2+2 \implies P(10)=1000+200+2=1202\), as you found.
Next, consider the case a=1 :
\(25b+5c+d=177 \implies b \le 7\), we see that, this is impossible. As, b is at most 5, giving 125, but c and d are forced to be 0. Hence, this case is rejected.
Proof:
\(25b \le 125, 5c \le 25, d \le 5 \implies 25b+5c+d \le 155\), which again, is not attainable anyway.
Therefore,\(\text{ }P(x)=x^3+2x^2+2\) is the only possible polynomial with the given conditions, and 1202 is the only valid solution.
I hope this helps!
