find the least square solution for the best parabola find the least squares solution for the best parabola going through (1,1), (2,1), (3,2), (4,2)

so to solve this problem I have 4 equation set up

  a +  b + c = 1
 4a + 2b + c = 1
 9a + 3b + c = 2
16a + 4b + c = 2


I found my $A$ to be 

1  1  1
4  2  1
9  3  1
16 4  1


and my $\mathbf{b}$ to be

1
1
2
2


then I apply the normal equation $x =(A^TA)^{-1} \, A^T b$ to solve for least square solutions 

I got $\mathbf{x}$ = 

0
0.4
0.5


which give me $a = 0, b = 0.4, c = 0.5$

my friend went through similar process and get $a = 0.5, b= 0.4, c = 0$

which should be the right solutions for this problem ? 

was it me that getting the reverse answer? or was my answer correct?

thank in advance
Most likely, you set up the quadratic as $ax^2+bx+c$, and your buddy set it up as $a+bx+cx^2$. Either way is OK, so long as when you get to the end you both remember which way you set up your quadratic.
Just a small note, "the best parabola" may be ambiguous term.* If the linear regression is specified in the task, it's clear. However, the metric of squares would suit me more to take the Euclidean distance of the points from parabola. See the code:

%% given points
x = [1 2 3 4]; 
y = [1 1 2 2];
N = length(x);

%% optimisation function 1
S1 = @(a) 0;
for k1 = 1:N
 S1 = @(a) S1(a) + dist2(x(k1),y(k1),a);
end

%% optimisation function 2
S2 = @(a) 0;
for k1 = 1:N
 S2 = @(a) S2(a) + (y(k1) - fun(x(k1),a(1),a(2),a(3)) ).^2;
end

a1 = fminsearch(S1,[0 0 0]);
a2 = fminsearch(S2,[0 0 0]); 
%% compare optimising functions
f1S1=S1(a1)
f1S2=S1(a2)
f1S2=S2(a1)
f2S2=S2(a2)

%% plot
xgrid = linspace(0,10,100);
plot(xgrid,fun(xgrid,a1(1),a1(2),a1(3))); hold on;
plot(xgrid,fun(xgrid,a2(1),a2(2),a2(3))); hold on;
scatter(x,y);

%% functions
function f=dist2(x,y,a)
 g = @(t) (x-t).^2 + (y-fun(t,a(1),a(2),a(3))).^2;
 f = fminsearch(g,0);
 f = g(f);
end

function f=fun(x,a,b,c)
 f = a*x.^2 + b*x +c;
end


In this case, I would even go further to use https://en.wikipedia.org/wiki/Parabola#General_case , but there are 5 parameters in a general parabola and only four points. So, there would be solutions including all the points on the parabola.

*And this nuance is definitely not just a mathematical prudence. Imagine for example, that you measured points in a 2D space and you know that your physical trajectory is a parabola. in this case the weight of both coordinates x and y is completely equivalent. I am almost sure that it's not OP's intent in this case. However in these kind of problems, where we usually analyse measured data, it may be good to keep in mind.