show that n 2+n+1 is not divisible by 5 for any n where n is a natural number
Answers
Any natural number it's "
n = 5k
n = 5k±1
n = 5k±2
where k natural number k
Case n = 5k then
n^2+n+1 = 25k^2+5k+1 = 5k(k+1)+1 then not divisible by 5 (remainder 1)
CAse n = 5k±1 then
n^2+n+1 = 25k^2 ± 10k+1 + 5k±1+1 = 5k(5k±1) +2±1 then not divisible by 5 (remainder 3 or 1)
CAse n = 5k±2
n^2+n+1 = 25k^2 ± 10k+4 + 5k±2+1 = 5k(5k±1+1) ± 2 then not divisible by 5 (remainder 2 or 3)
Then
n^2+n+1 not divisible by 5
B
If n^2+n+1 divisible by 5 then
exist k like
n^2+n+1=5k then
n^2+n+1-5k=0
n zeroes (root) of n^2+n+1-5k=0
If n natural first n must be real then integer then positive
Discriminant =
D= sqrt(1-4+20k)=sqrt(20k-3)
20k-3 must be perfect square let be m^2 then
20k-3 = m^2
k = (1/20)(m^2+3)
m^2 +3 must dividing by 10 then m^2 must finishing in ...7
No perfect square finishing in 7 then no m no k then n it's irational then no exist natural n
5∤(n2+n+1) for all integer n.
Because n2+n+1≡(n+3)2+2(mod5)
So if 5∣(n2+n+1) for some n, then n satisfies (n+3)2≡3(mod5). But 3 is not quadratic residue modulo 5. (You can check that 3 is not quadratic residue easily.)
If you don't know modular arithmetic you can also show it by induction.
Prove that if n2+n+1 is not divisible by 5, then also (n+5)2+(n+5)+1 is not divisible by 5.
Prove explicitly the base cases: n=1,n=2,n=3,n=4,n=5.
By induction this then holds for all n.