# Understand the problem

Consider the following \( 3 \times 2 \) array formed by using the numbers \( 1 , 2 , 3 ,4 ,5 , 6 \ : \)
\( \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} = \begin{pmatrix} 1 &6 \\ 2 & 5 \\ 3 & 4 \end{pmatrix} \) . Observe that all row sums are equal , but the sum of the squares is not the same for each row .Extend the above arrar to a \( 3 \times k \) array \( (a_{ij}){3 \times k} \) for a suitable k , adding more columns , using the numbers \( 7 , 8 , 9 , ….,3k \) such that \( \displaystyle \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} \ and \ \sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 \)

##### Source of the problem

Regional Math Olympiad, 2019 Problem 4

##### Topic

Inequality

##### Difficulty Level

8/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics

# Start with hints

Do you really need a hint? Try it first!

1+2+3+ … 3k = \( \frac{3k(3k+1)}{2}\) So, \( \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} = \frac{k(3k+1)}{2}\) \( 1^2 + 2^2 +… (3k)^2 = \frac{k(3k+1)(6k+1)}{2}\) So, \(\sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 = \frac{k(3k+1)(6k+1)}{6}\). This means that 3 | k.

Step 1: Try out with k = 3. Prove that it is not possible to arrange them in the desired order as already some numbers are fixed.
We will now try for k = 6.
Claim: If 3|k and k > 3 then it is always possible. (This is our conjecture too)

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5)
\(n^2 + (n + 5)^2 – (n + 1)^2 – (n + 4)^2 = 8\)
\( (m + 1)^2 + (m + 4)^2 – (m + 2)^2 – (m + 3)^2 = 4\)
\( (l + 2)^2 + (l + 3)^2 – (l)^2 – (l+ 5)^2 = – 12 \)
So, we get \( n^2 + (n + 5)^2 + (m + 1)^2 + (m + 4)^2 + (l + 2)^2 + (l + 3)^2 = (n + 1)^2 + (n + 4)^2 + (m + 2)^2 +(m + 3)^2 + (l)^2 + (l+ 5)^2 \).
Also,

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5) Hence putting suitable values of l, m, and n, we get an array like the one below: \( \begin{pmatrix} 1 & 6 & 8 & 11 & 15 & 16 \\ 2 & 5 & 9 & 10 & 13 & 18 \\ 3 & 4 & 7 & 12 & 14 & 17\end{pmatrix} \) \( (n + 1)^2 + (n + 6)^2 + (n + 8)^2 + (n + 11)^2 + (n + 15)^2 + (n + 16)^2

= (n + 2)^2 + (n + 5)^2+ (n + 9)^2 + (n + 10)^2+ (n + 13)^2+ (n + 18)^2

= (n + 3)^2 + (n + 4)^2 + (n + 7)^2+ (n + 12)^2 + (n+14)^2 + (n+17)^2 \). Using this and the above two matrices, you can prove by induction that the claim holds!

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5) Hence putting suitable values of l, m, and n, we get an array like the one below: \( \begin{pmatrix} 1 & 6 & 8 & 11 & 15 & 16 \\ 2 & 5 & 9 & 10 & 13 & 18 \\ 3 & 4 & 7 & 12 & 14 & 17\end{pmatrix} \) \( (n + 1)^2 + (n + 6)^2 + (n + 8)^2 + (n + 11)^2 + (n + 15)^2 + (n + 16)^2

= (n + 2)^2 + (n + 5)^2+ (n + 9)^2 + (n + 10)^2+ (n + 13)^2+ (n + 18)^2

= (n + 3)^2 + (n + 4)^2 + (n + 7)^2+ (n + 12)^2 + (n+14)^2 + (n+17)^2 \). Using this and the above two matrices, you can prove by induction that the claim holds!

Using the array found similarly :
\( \begin{pmatrix} 1 & 6 & 8 & 11 & 18 & 13 & 21 & 23 & 25 \\ 2 & 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27 \\ 3 & 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{pmatrix} \)

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