Question

what is this?!

Answer 1

can you help me with my math question?

Answer 2

this is sparta xD

Answer 3

Prove that there always exist three numbers \(a,b,c\) from any given seven integers such that \(a^2 + b^2 + c^2 - ab - bc - ca\) is divisible by 7.

Answer 4

consecutive even integers ?

Answer 5

\[= \frac{1}{2} \left((a-b)^2 + (b-c)^2 + (c-a)^2\right)\]if \(a,b , c\) are equivalent modulo 7 then the claim is kinda trivial

Answer 6

yes

Answer 7

for ever positive variable^2 u have a negative version, which will eliminate the remainders

Answer 8

and it will be div by 2, as u get 2 instances of mod 7 integers being added together

Answer 9

ok so since we know that the whole expression is an integer we can maybe remove the 1/2 part and concentrate on the expression itself

if a, b, are equivalent modulo 7 then (a-b)^2 is divisible by 7 by definition while \((b-c)^2 + (a-c )^2=b^2 + a^2 - 2bc - 2ac + 2c^2 \)

Answer 10

why are u saying (a-b)^2 is div by 7

Answer 11

a,b,c all dont have to be div by 7, but when considered in mod 7, ull see that for all their positive remainder residues, u have a term subtracting them off again

Answer 12

if a, b are equivalent modulo 7

Answer 13

i guess i dont know what that means

Answer 14

a=bmod7?

Answer 15

we dont need to worry about and equivalence we can solve it straight ahead

Answer 16

yes

Answer 17

with the explaination i gave a bit earlier