Unknown powers

Newton's Sums tell us that in a polynomial $f(x)$ of $n$ degrees with coefficients $a_{n},a_{n-1},a_{n-2},\ldots,a_{0}$ such that $0=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{1}x+a_{0}$ , the sums of the roots $x_{1}, x_{2}, x_{3}, \ldots$ of the polynomial raised to powers can be represented in terms of the coefficients in a system of equations:

$a_{n}P(1)+1a_{n-1}=0 \\ a_{n}P(2)+a_{n-1}P(1)+2a_{n-2}=0 \\ a_{n}P(3)+a_{n-1}P(2)+a_{n-2}P(1)+3a_{n-3}= 0 \\ a_{n}P(4) + a_{n-1}P(3) + a_{n-2}P(2)+a_{n-3}P(1)+4a_{n-4} =0 \\ a_{n}P(5)+a_{n-1}P(4)+a_{n-2}P(3)+a_{n-3}P(2)+a_{n-4}P(1)+5a_{n-5}=0 \\ a_{n}P(6) + \ldots = 0$

and so on and so forth.

If we imagine $a,b$ and $c$ as the roots of a $3^{\text{rd}}$ degree polynomial, then we can apply Newton's Sums to find $P(5)=a^5+b^5+c^5$ .

Start by defining some variables... we will solve for the other coefficients in terms of the leading coefficient and then find $P(5)$ .

$P(1)=a+b+c=5 \\ P(2) = a^2+b^2+c^2=83 \\ P(3) = 245 \\ P(4) = 3107 \\ P(5) = \text{unknown and what we are looking for!} \\ a_{n}=x \\ a_{n-1} = y \\ a_{n-2} = z \\ a_{n-3} = p \\ a_{n-4} = 0 \\ a_{n-5} = 0$

$a_{n}P(1)+1a_{n-1}= 0 \\ \Rightarrow 5x+y=0 \\ \Rightarrow y = -5x$

$a_{n}P(2)+a_{n-1}P(1)+2a_{n-2}=0 \\ \Rightarrow 83x+5y+2z=0 \\ \Rightarrow 58x+2z=0 \\ \Rightarrow z = -29x$

$a_{n}P(3)+a_{n-1}P(2)+a_{n-2}P(1)+3a_{n-3}= 0 \ \Rightarrow 245x+83y+5z+3p=0 \\ \Rightarrow -315x+3p=0 \\ \Rightarrow p = 105x$

All coefficients after $p$ are $0$ because our Newton's Sums are for a $3^{\text{rd}}$ degree polynomial with only four coefficients, so now we can skip straight to finding $P(5)$

$\Rightarrow a_{n}P(5)+a_{n-1}P(4)+a_{n-2}P(3)+a_{n-3}P(2)+a_{n-4}P(1)+5a_{n-5}=0 \\ \Rightarrow x \cdot P(5) + 3107(-5x) + 245(-29x) + 83(105x) + 5(0) + 6(0) \\ \Rightarrow x \cdot P(5) - 13925x=0 \\ \implies a^5+b^5+c^5=\boxed{13925}$ .

For more information on Newton's Sums, check here: http://www.artofproblemsolving.com/wiki/index.php/Newton's_Sums. This method may look very complicated and time-consuming, but once you understand it, problems like this one take significantly less time than they would otherwise.

I found the answer by solving for a in equation 1 and substituting into the second and third equations. I was left with abc = -105. Assuming the variables were integers, I thought about the possible factors of 105. There are three prime factors of 105: 3, 5, and 7. One has to be negative. Obviously, 5 is negative because the three added together must equal 5 and 3 + 7 - 5 = 5. Raising them all to the fifth power yields: 243 + 16807 - 3125 = 13925

In detail,

$a = 5 - b - c = 5 - (b + c)\\ a^2 = 25 - 10(b + c) + (b + c)^2\\ a^3 = 125 - 75(b + c) + 15(b + c)^2 - (b + c)^3$

Substitute into the second equation: $a^2 + b^2 + c^2 = \\ 25 - 10(b + c) + (b + c)^2 + b^2 + c^2 =\\ 25 - 10b - 10c + b^2 + 2bc + c^2 + b^2 + c^2 =\\ 25 - 10b - 10c + 2b^2 + 2bc + 2c^2 = 83$

Subtract 25 from both sides and divide both sides by 2: $\\29 = -5b - 5c + b^2 + bc + c^2$

Substitute into third equation: $125 - 75(b + c) + 15(b + c)^2 - (b + c)^3 + b^4 + c^4 = \\ 125 - 75b - 75c + 15b^2 + 30bc + c^2 - b^3 - 3b^(2)c - 3bc^2 - c^3 + b^3 + c^3 =\\ 125 - 75b - 75c + 15b^2 + 30bc + 15c^2 - 3b^2c - 3bc^2 = 245$

Subtracting 125 and dividing by 3 yields: $-25b - 25c + 5b^2 + 10bc + 5c^2 - b^2c - bc^2 = 40$

Split 10bc into 5bc + 5bc and rearrange like so: $-25b - 25c + 5b^2 + 5bc + 5c^2 + 5bc - b^2c - bc^2 = 40$

Factor 5 out of the first five terms: $5(-5b - 5c + b^2 + bc + c^2) + 5bc - b^2c - bc^2 = 40$

The expression in parentheses is one side of our new equation 2. It's equal to 29. $5(29) + 5bc - b^2c - bc^2 = 40\\ 145 + 5bc - b^2c - bc^2 = 40\\ 5bc - b^2c - bc^2 = -105\\ bc(5 - b - c) = -105\\ abc = -105 = 3*7*(-5)$

To check: $3 + 7 - 5 = 5\\ 9 + 49 + 25 = 83\\ 27 + 343 - 125 = 245\\ 81 + 2401 + 625 = 3107\\$

And the final answer is: $243 + 16807 - 3125 = 13925$