2017 JBMO Problems/Problem 1
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
Every set which a solution is of the form 
Since they are consecutive, it follows that 
are even and 
are odd.
In addition, each pair of positive integers destined to be multiplied together can have a difference of either 
or 
or 
.
So, we only have to consider integers from up to 
. Therefore we calculate the following products:
A = { 2⋅3, 4⋅5, 6⋅7, 8⋅9 10⋅11, 11⋅12}
B = { 2⋅1, 4⋅3, 6⋅5, 8⋅7 10⋅9, 11⋅13}
C = { 1⋅4, 2⋅5, 3⋅6, 4⋅7 5⋅8, 6⋅9, 7⋅10, 8⋅11, 9⋅12}
D = { 1⋅6, 2⋅7, 3⋅8, 4⋅9 5⋅10, 6⋅11, 7⋅12}
And we can only have five cases:
Case 1: 
Y_2 is the only solution set for this case.
Case 2: 
Y_1 is the only solution set for this case.
Case 3: 
No solution set for this case.
Case 4: 
No solution set for this case.
Case 5: 
No solution set for this case.
See also
| 2017 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
