## Tuesday, December 7, 2010

### a,b,c,d odd powers

If $a,b,c,d$ are integers such that $a+b+c+d = 0$ then show that for any odd number $n$,

$$X_n = a^n + b^n + c^n + d^n$$

is divisible by $Y = \sqrt{-(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)}$

And that if $n$ is prime, then $X_n$ is divisible by $n$ as well

First Solution

First, it is straightforward to show that $Y$ is indeed an integer.
$a+b = -(c+d)$
$a+c = -(b+d)$
$a+d = -(b+c)$
Multiplying all three, we have that $Y = |(a+b)(b+c)(c+a)|$
If $Y=0$, then $X_n = 0$ as well. So now we're going to assume that $Y > 0$. For proving divisibility purposes, we can also assume that $Y = (a+b)(b+c)(c+a)$.

Let $f(t) = (t-a)(t-b)(t-c) = t^3 - Pt^2 + Qt - R$ be a polynomial with roots $a,b,c$

We have that $Y=(a+b)(b+c)(c+a) = (P-a)(P-b)(P-c) = f(P) = PQ-R$.

Let's also define $S_n = a^n + b^n + c^n$. All we have to show is that $S_n - P^n$ is divisible by $Y$ for $n$ odd.

Now we assert the following, and prove by induction:
If $n$ is odd, then $S_n - P^n$ is divisible by $Y$
If $n=2m$ is even, then $S_n - P^n - 2(-Q)^m$ is divisible by $Y$.

We readily obtain that $S_1 = P$ and $S_2 = P^2 - 2Q$, so the assertions above are true for $n=1,2$.

Now for higher values of $n$, we have the following:
$a^{n-3}f(a) = a^n - Pa^{n-1} + Qa^{n-2} - Ra^{n-3} = 0$
similarly we can do the same for $b,c$ and add them, so that we have:
$S_n = PS_{n-1} - QS_{n-2} + RS_{n-3}$

If $n=2m$ is even, then $n-1$ is odd, $n-2 = 2(m-1)$ is even, and $n-3$ is odd. So from induction hypothesis, we can write:
$S_{n-1} = P^{n-1} + YZ_1$
$S_{n-2} = P^{n-2} + 2(-Q)^{m-1} + YZ_2$
$S_{n-3} = P^{n-3} + YZ_3$

where $Z_1, Z_2, Z_3$ are integers
So:
$$S_n = P^n + PYZ_1 - P^{n-2}Q + 2(-Q)^m -QYZ_2 + RP^{n-3} + RYZ_3$$
$$= P^n + 2(-Q)^m + P^{n-3}(PQ-R) + Y(PZ_1 - QZ_2 + RZ_3)$$
$$= P^n + 2(-Q)^m + Y(P^{n-3} + PZ_1 - QZ_2 + RZ_3)$$

which proves our induction assertion that $S_n - P^n - 2(-Q)^m$ is divisible by $Y$. For $n$ odd, we can do the same.

So $S_n - P^n = X_n$ is divisible by $Y$ for $n$ odd.

Second Solution

We have $d = -(a+b+c)$, then
$$X_n = a^n + b^n + c^n - (a+b+c)^n$$ if $n$ is odd.

For fixed values of $b,c$, let $f(a) = a^n + b^n + c^n - (a+b+c)^n$ by a polynomial of degree $n-1$ in $a$.
Note that $f(-b) = 0$ and $f(-c) = 0$. That means $X_n = f(a)$ is divisible by $(a+b)(a+c)$.

Write $f(a) = (a+b)(a+c)P(a)$ for $P$ some integer polynomial.
Now note that $f(a) = a^n - (a+b+c)^n + (b^n + c^n)$
$a^n - (a+b+c)^n$ is divisible by $(b+c)$ in a polynomic manner (not just integer divisibility), and so is $b^n + c^n$ (since $n$ is odd).

So $f(a)$ is divisible by $(b+c)$ in a polynomic manner, which means $P(a)$ is divisible by $(b+c)$.