- #1
Detektyw
- 6
- 0
Homework Statement
Let's take a prime number p not equal to 5.
Now let's take three integers a,b,c.
Prove that if p | (a + b + c) [tex]\wedge[/tex] p | (a^5 + b^5 + c^5), then
p | (a^2 + b^2 + c^2) [tex]\vee[/tex] p | (a^3 + b^3 + c^3)
Homework Equations
I think:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
and
the same for third and fifth power.
The Attempt at a Solution
If either (a^2 + b^2 + c^2) or (a^3 + b^3 + c^3) is divisible by p, then the product of them will too be divisible by p.
(a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) = (a + b + c)^5 - 3(a + b + c)^2 * (ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2 + 2ab) - 2(a + b + c)^3 * (ab + bc + ca) + 6(((a + b + c)^2 - (a^2 + b^2 + c^2)) * ((a + b + c)^3 - (a^3 + b^3 - c^3)))
(a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) = (a + b + c)^5 - 3(a + b + c)^2 * (ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2 + 2ab) - 2(a + b + c)^3 * (ab + bc + ca) + 6((a + b +c)^5 - (a + b + c)^2 * (a^3 + b^3 + c^3) - (a + b + c)^3 * (a^2 + b^2 + c^2) + (a^2 + b^2 + c^2) * (a^3 + b^3 + c^3))
By moving 6 * (a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) on the right hand side:
-5 * (a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) = 7 * (a + b + c)^5 - 3 * (a + b + c)^2 * (ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2 + 2ab) - 2 * (a + b + c)^3 * (ab + bc + ca) - 6 * (a^2 + b^2 + c^2) * (a + b + c)^3 - 6 * (a^3 + b^3 + c^3) * (a + b + c)^2
In a different arrangement:
-5 * (a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) = (a + b + c)^2 * (7 * (a + b + c)^3 - 3 * (ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2 + 2ab) - 2 * (a + b + c) * (ab + bc + ca) - 6 * (a^2 + b^2 + c^2) * (a + b + c) - 6 * (a^3 + b^3 + c^3))
Which basically proves that 5 * (a^2 + b^2 + c^2) * (a^3 + b^3 + c^3) is divisible by p, as every term on the right hand side is a product of (a + b + c) and something else, which by the assumptions is divisible by p.
It seems that I'd have to figure out some way to prove that either (a + b + c)^2 is divisible by 5, or that 7 * (a + b + c)^3 - 3 * (ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2 + 2ab) - 2 * (a + b + c) * (ab + bc + ca) - 6 * (a^2 + b^2 + c^2) * (a + b + c) - 6 * (a^3 + b^3 + c^3) is divisible by 5, as either of these would prove the thesis.
Comments and hopefully pointers appreciated.
Thanks in advance,
Detektyw