As $(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$ satisfy the following:
$\alpha_m= am^2 +bm +c$ for $m=1,2,3,4$ for some real numbers $a,b,c.$
When we consider a 4 x 4 square grid and fill it such that the ordered quadruples making up all 4 rows as well as the ordered quadruples making up the first 3 columns.
In order to satisfy the above condition, lets see the relations between the $\alpha_i$ of each column given that we have 4 ordered quadruples in each row.
The $c_i$ must be equal for all the $\alpha_i$ in the grid else the constraint will immediately be violated.
The $b_i$ must have the ratio 1:2:3:4 for each column assuming The $b_i$ must have the ratio 1:2:3:4 for each column. (Substitute and try it out to get the feel)
The $a_i$ must have the ratio 1:4:9:16 for each column so that all the columns satisfy the condition of fitting in the grid. ($a_i$ is the quadratic coefficient, hence the ratio for $b_i$ will be squared)
Finally, lets consider the last column:
we have (16$a_1$ + 4$b_1$ + $c_1$, 16$a_2$ + 4$b_2$ + $c_2$, 16$a_3$ + 4$b_3$ + $c_3$, 16$a_4$ + 4$b_4$ + $c_4$)
As all the $c_i$ are equal, substituting the ratios from earlier, we get:
16$a_1$ + 4$b_1$ + $c_1$, 64$a_1$ + 8$b_1$ + $c_1$, 144$a_1$ + 12$b_1$ + $c_1$, 256$a_1$ + 16$b_1$ + $c_1$
which follow the ordered quadruple $\alpha_m= 16a_1m^2 +4b_1m +c_1$ for $m=1,2,3,4$