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Moderator Note: This question is from a contest which ended on 22 Oct 2012.

Consider $(\alpha_1, \alpha_2, \alpha_3, \alpha_4)$ such that the ordered quadruple satisfies the following:

$\alpha_m= am^2 +bm +c$ for $m=1,2,3,4$ for some real numbers $a,b,c.$

Suppose we consider a square grid that measured four on each side, and fill it such that the ordered quadruples making up all the rows as well as the ordered quadruples making up the first three columns satisfied the above condition. How can we prove that the fourth column will also contain an ordered quadruple satisfying the condition?

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A sequence $(x_k)_{k\geq 1}$ of numbers can be produced by a linear polynomial in the form $x_k= bk + c\qquad(k\geq 1)$ with suitable coefficients $b$, $c$ iff the first differences $x_{k+1}-x_k$ are all equal. Similarly, the $x_k$ can be produced by a quadratic polynomial in the form $x_k=a k^2 + bk + c\qquad(k\geq1)$ with suitable coefficients $a$, $b$, $c$ iff their second differences $(x_{k+2}-x_{k+1})-(x_{k+1}-x_k)$ are all equal.

In our case of just four numbers $x_1$, $x_2$, $x_3$, $x_4$ this amounts to the single condition $(x_4-x_3)-(x_3-x_2)=(x_3-x_2)-(x_2-x_1)\ ,$ which is the same as $x_4=3x_3-3x_2+x_1\ .\qquad(*)$ Using this for the rows of your$4\times4$ matrix (not "grid") $A=\bigl[a_{ik}\bigr]$ we see that it is of the form $\bigl[a_{ik}\bigr]=\left[\matrix{ a_{11}&a_{12}&a_{13}& (3a_{13}-3a_{12}+a_{11}) \cr a_{21}&a_{22}&a_{23}& (3a_{23}-3a_{22}+a_{21}) \cr a_{31}&a_{32}&a_{33}& (3a_{33}-3a_{32}+a_{31}) \cr a_{41}&a_{42}&a_{43}& (3a_{43}-3a_{42}+a_{41}) \cr}\right]\ .$ Since the first three columns of $A$ should also fall into this pattern we necessarily have $a_{41}=3 a_{31}-3a_{21}+a_{11}\ ,\quad a_{42}=3 a_{32}-3a_{22}+a_{12}\ , \quad a_{43}=3 a_{33}-3a_{23}+a_{13}\ .$ This implies $\eqalign{a_{44}&= 3a_{43}-3a_{42}+a_{41}\cr &=3(3 a_{33}-3a_{23}+a_{13})-3(3 a_{32}-3a_{22}+a_{12})+(3 a_{31}-3a_{21}+a_{11})\cr & =3(3a_{33}-3a_{32}+a_{31})-3(3a_{23}-3a_{22}+a_{21})+(3a_{13}-3a_{12}+a_{11})\cr &=3a_{34}-3 a_{24}+a_{14}\ .\cr}$ This shows that the fourth column of $A$ fulfills condition $(*)$ automatically.

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    @Danielle Huang: The occurring terms have been grouped in$a$different way. There is no reversing of indices here.2012-10-22
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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 $m=1,2,3,4$

QED

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    Similarly, for $b_i$, lets compare the 1st two $\alpha_i$ in the first column. We have $\alpha_11$ = a11 + b11 + c1. $\alpha_12$ = a12 + b12 + c12. For both these to be part of the *same* quadruple (given that the 1st column is a quadruple), $a_12$ = 4$a_11$, $b_12$ = 2$b_11$ (as m = 2)2012-09-20