# 2017 JBMO Problems/Problem 1

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## Problem

Determine all the sets of six consecutive positive integers such that the product of some two of them, added to the product of some other two of them is equal to the product of the remaining two numbers.

## Solution

$x_1 x_2 + x_3 x_4 = x_5 x_6$

Every set which is a solution must be of the form $Y_k = \{k, k+1, k+2, k+3, k+4, k+5\}$

Since they are consecutive, it follows that $x_2, x_4, x_6$ are even and $x_1, x_3, x_5$ are odd.

In addition, exactly two of the six integers are multiples of $3$ and need to be multiplied together. Exactly one of these two integers is even (and also the only one which is multiple of $6$) and the other one is odd.

Also, each pair of positive integers destined to be multiplied together can have a difference of either $1$ or $3$ or $5$.

So, we only have to consider integers from $1$ up to $11$ since $k \leq 6$ (see inequalities below). Therefore we calculate the following products:

A = { 2⋅1, 4⋅5, 8⋅7, 13⋅14, 10⋅11}

B = { 1⋅4, 2⋅5, 3⋅6, 4⋅7, 5⋅8, 6⋅9, 7⋅10, 8⋅11}

C = { 2⋅7, 5⋅10}

In any case, either 3⋅6 or 6⋅9 needs to be included in every solution set. {7⋅10, 8⋅11} cannot be part of any potential solution set, but they were included here just for completeness.

Method 1: Using a graph

One could also construct a graph G=(V,E) with the set V of vertices (also called nodes or points) and the set E of edges (also called arcs or line). The elements of all sets A,B,C will be the vertices. The edges will be the possibles combinations, so the candidate solutions will form a cycle of exactly three vertices. So, there should be eight such cycles. Only three of them will be the valid solution sets:

$2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$

$1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5$

$7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11$

Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}, { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}, {3⋅6, 2⋅7, 4⋅5}

Method 2: Taking cases

Alternatively, we can have five cases at most (actually only three) :

Case 1: $|x_1-x_2| = |x_3 -x_4| = |x_5 -x_6| = 3$

$k(k+3)+(k+1)(k+4) \leq (k+2)(k+5) \to k \leq 3$

We just have to look at set B in this case.

$2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$

Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}

$Y_2$ is the only solution set for this case.

Case 2: $|x_1-x_2| =1, |x_3 -x_4| = 3, |x_5 -x_6| = 1$

$k(k+1)+(k+2)(k+5) \leq (k+3)(k+4) \to k \leq 3$

$k(k+3)+(k+1)(k+2) \leq (k+4)(k+5) \to k \leq 6$

$1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5$

$7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11$

Rejected: { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}

$Y_1$ and $Y_6$ are the only solution sets for this case.

Case 3: $|x_1-x_2| = 3 |x_3 -x_4| = 5, |x_5 -x_6| = 1$

$k(k+5)+(k+1)(k+4) \leq (k+3)(k+5) \to k \leq 2$

Rejected: {3⋅6, 2⋅7, 4⋅5}

No solution set for this case since they were all rejected.

Case 4: $|x_1-x_2| = |x_3 -x_4| = |x_5 -x_6| = 1$

No solution set for this case, as the multiples of three need to be multiplied together.

(This case is actually not realistic and it was included here just for completeness.)

Case 5: $|x_1-x_2| = 1, |x_3 -x_4| = 5, |x_5 -x_6| = 1$

No solution set for this case, as the multiples of three need to be multiplied together.

(This case is actually not realistic and it was included here just for completeness.)

# Solution 2

Let the six numbers be $a, a+1, a+2, a+3, a+4, a+5$. We can bound the graph to restrict the values of $a$ by setting the inequality $(a+5)(a+4)\geq a(a+3)+(a+2)(a+1)$. If we sum the pairwise products, we show that the minimum sum obtainable is $a(a+3)+(a+2)(a+1)$ by first establishing that the $a^2$ and $a$ terms will be the same no matter how you pair them. However, we can manipulate the products such that we obtain the least possible constant, which is achievable by treating $a=a+0$ so we remove the greatest constant factor which is 3. Solving the inequality and making the left side 0, we get $0\geq a^2-6a-18=(a+3)(a-6)$. If the quadratic is greater than 0, it means that the RHS will be too large for any $a$ to suffice because we have already chosen the minimum value for the RHS and the maximum for the LHS. Clearly, $a=6$ works as a set. Now, we find the second largest product for the LHS and the smallest pairwise product-sum for the RHS. This is achievable at $(a+5)(a+3)\geq a(a+4)+(a+2)(a+1) \Longrightarrow 0\geq a^2-a-13$. Solving for the quadratic, we see that there are no integer roots for $a$ but we can bound $a\leq 4$. Now, we take the third largest product for the LHS which is $(a+4)(a+3)\geq a(a+5)+(a+2)(a+1) \Longrightarrow 0\geq a^2+a-10$ bounding $a\leq 2$. Now, we can further bound(simply use the method above) to see no solutions here.