It was much easier to use the first equation to give b= -a²-a, sub this into the other equation leads to 2a^4-a²-1=0. Which is a hidden quadratic that leads to a=1 or -1, which leads to b =0 or -2.
This looks like a pain to go trough normally. I think i got some ideas Since x^-ax+a^4 = q(x)(x+b)+1 X^2-ax+a^4-1= q(x)(x+b) plug in x=-b b^2+ab+a^4-1=0 The s cond statement makes 1/a root of the polynomial given so: b/a^2+1/a+1=0 b+a+a^2=0 so b=-a-a^2 Verifying in the first relation: A^4+2a^3+a^2-a^2-a^3+a^4-1=0 2a^4+a^3-1=0 a^4. a^3. a^2. a. a^0 Coef. 2. 1. 0. 0. -1 -1. 2(-1)+1. 2 -1. 1. -1. 0 P(a)=(a-1)(2a^3-a^2+a-1) R'(x)= 6a^2-2a+1 has no real roots So there are only 2 values for a. a=1 b=0 2a^3-a^2+a-1=0 2a(a^2+a-a)-a^2+a-1=0 -2ab -3a^2-3a+3a+a-1=0 -2ab+3b+4a-1=0 B=(1-4a)/(3-2a) so there are total 2 values of b one of which is 0
It was much easier to use the first equation to give b= -a²-a, sub this into the other equation leads to 2a^4-a²-1=0. Which is a hidden quadratic that leads to a=1 or -1, which leads to b =0 or -2.
I did it exactly the same way
Same!
Yeah I saw this solution after recording and was surprised I didn’t get it that way on the first go.
@@mathoutloud Haha, I'm often surprised myself at how “complicated” I solve various math problems, even though it would be much easier ;)
This looks like a pain to go trough normally. I think i got some ideas
Since x^-ax+a^4 = q(x)(x+b)+1
X^2-ax+a^4-1= q(x)(x+b) plug in x=-b
b^2+ab+a^4-1=0
The s cond statement makes 1/a root of the polynomial given so:
b/a^2+1/a+1=0
b+a+a^2=0 so b=-a-a^2
Verifying in the first relation:
A^4+2a^3+a^2-a^2-a^3+a^4-1=0
2a^4+a^3-1=0
a^4. a^3. a^2. a. a^0
Coef. 2. 1. 0. 0. -1
-1. 2(-1)+1.
2 -1. 1. -1. 0
P(a)=(a-1)(2a^3-a^2+a-1)
R'(x)= 6a^2-2a+1 has no real roots
So there are only 2 values for a. a=1 b=0
2a^3-a^2+a-1=0
2a(a^2+a-a)-a^2+a-1=0
-2ab -3a^2-3a+3a+a-1=0
-2ab+3b+4a-1=0
B=(1-4a)/(3-2a) so there are total 2 values of b one of which is 0