2023/034) Given $a+b+c=3$ and $a^2+b^2+c^2 = 3$ Solve for real a,b,c

We have 

$a+b+c=3\cdots(1)$

$a^2+b^2+c^2=3\cdots(2)$

Squaring (1) we get

$a^2+b^2 + c^2 + 2ab + 2bc + 2ca = 9$

Putting the value of $a^2+b^2+c^2$ from (2) we get

$ab+bc+ca = 3$

subracting above from (2) we get

$a^2+b^2+c^2 - ab - bc - ca = 0$

or $2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0$

or $(a^2-2ab + b^2) + (b^2-2bc+c^2) + (c^-2ca +a^2) = 0$

or $(a-b)^2 + (b-c)^2 + (c-a)^2= 0$

above is possible only if a=b= c

putting in (1) we get a = b = c = 1