(1+a+b)2=3(1+a2+b2) Let's do it???

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2 Answers
Cesareo R.
Aug 11, 2017

a=1,b=1

Explanation:

Solving the traditional way

(1+a+b)2−3(1+a2+b2)=0⇒1−a+a2−b−ab+b2=0

Now solving for a

a=12(1+b±√3√2b−b2−1) but a must be real so the condition is

2b−b2−1≥0 or b2−2b+1≤0⇒b=1

now substituting and solving for a

1−2a+a2=0⇒a=1 and the solution is

a=1,b=1

Another way to do the same

(1+a+b)2−3(1+a2+b2)=0⇒1−a+a2−b−ab+b2=0

but

1−a+a2−b−ab+b2=(a−1)2+(b−1)2−(a−1)(b−1)

and concluding

(a−1)2+(b−1)2−(a−1)(b−1)=0⇒a=1,b=1

George C.
Aug 11, 2017

D. There is exactly one solution pair (a,b)=(1,1)

Explanation:

Given:

(1+a+b)2=3(1+a2+b2)

Note that we can make this into a nice symmetric homogeneous problem by generalising to:

(a+b+c)2=3(a2+b2+c2)

then set c=1 at the end.

Expanding both sides of this generalised problem, we have:

a2+b2+c2+2ab+2bc+2ca=3a2+3b2+3c2

Subtracting the left hand side from both sides, we get:

0=2a2+2b2+2c2−2ab−2bc−2ca

0=a2−2ab+b2+b2−2bc+c2+c2−2ca+a2

0=(a−b)2+(b−c)2+(c−a)2

For real values of a, b and c, this can only hold if all of (a−b), (b−c) and (c−a) are zero and hence:

a=b=c

Then putting c=1 we find the only solution to the original problem, namely (a,b)=(1,1)

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