Every nice maths olympiad algebra problem | Can you solve for a+b=? | Algebra

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick.
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[@damyankorena]

I am bored so lets solve with complex numbers.
4+4i=a²+2abi-b²=(a+bi)²
a+bi=√(4+4i)
Let z=4+4i
|z|=4√2
z/|z|= 1/√2 + i/√2
Arg(z)=π/4
Therefore
√z=2∜2 * e^(πi/8)
Which as shown earlier is equal to a+bi.
a=Re(√z), b=Im(√z)
a=2∜2 * sin(π/8)
b=2∜2 * cos(π/8)
a+b=2∜2 * √2 sin(⅜π)
a+b= 2^(¾) * √(2+√2)
а+b=2√(1+√2)
Naturally while taking the square root of z instead of
Arg(√z)=⅛π we can use
Arg(√z)=⅝π which will by similar methods turn out to be the other solution in which
a+b=-2√(1+√2)

[@ohiyo_o7637]

My approach is:
(a+b)(a-b)=4
(a^2+b^2+4)(a^2+b^2-4)=16
(a^2+b^2)^2=32
a^2+b^2=4root(2) (notice that a^2 and b^2 are both positive, so no plus or minus sign is needed for this square root)
(a+b)^2=4root(2)+4
a+b=plus or minus 2root((root(2)+1))

[@ohiyo_o7637]

i got a+b = root( root(32)+4 ), is it correct or where did i do it wrong

[@ohiyo_o7637]

oh it is the same thing, i just didnt simplify it, i am not very good at math haha

[@damyankorena]

√32=4√2
So
√(4+√32)=√(4+4√2)
= 2√(1+√2)
Therefore they are equal

[@ohiyo_o7637]

I see, thank you