mystic finch
After many hours on trying to prove x^2+10x-3 is irreducible over Q without using common methods, I've found that if x^2+10x-3 is reducible over Q, then
3(2k+1)^2-(2p+1)^2 must be a multiple of 10 and [3(2k+1)^2]/[2p+1] must be an integer. (2k+1) and (2p+1) must be co prime
Prove there is no k,p∈Z that satisfies all conditions
valid notchBOT
elfin locust
Can you show your work? I'm interested.
Here is another way to show irreducibility.
Let f(x)=x^2+10x-3.
Then f(x) is irreducible if and only if g(x)=f(x+12) is irreducible.
Moreover, g(10)=f(22)=701 is a prime and the degree of g is less than or equal to 31.
Therefore g(x) and thus f(x) is irreducible.
mystic finch
mystic finch
@sullen fable
limpid ridge
<@1177922913605795882>
2p+1 must divide 3
There are finitely many divisors of 3, so you should be able to just case bash it
limpid ridge
[3(2k+1)^2]/[2p+1]∈Z has solutions even if 2p+1 doesn't divide 3
Not if 2p+1 and 2k+1 are coprime
If a divides bc and a is coprime with c then a divides b
mystic finch
2(-3)+1=-5?
limpid ridge
2p+1 not p
limpid ridge
The possible values for p are -2, -1, 0 and 1
mystic finch
If 2p+1 and 2k+1 are coprime then 2p+1 can't be 1 or -1 right?