2021 AIME I Problems/Problem 5
Contents
Problem
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
Solution 1
Let the terms be , , and . Then we want , or . Rearranging, we get . Simplifying further, . Looking at this second equation, since the right side must be an integer, must equal . Looking at the first equation, we see since is positive. This means we must test . After testing these, we see that only and work which give and respectively. Thus the answer is . ~JHawk0224
Note: If you don't understand the simplification of , you an actually just use synthetic division and arrive at the same place ~ Anonymous
Solution 2
Let the common difference be and let the middle term be . Then, we have that the sequence is This means that the sum of the squares of the 3 terms of the sequence is We know that this must be equal to so we can write that and it follows that
Now, we can treat as a constant and use the quadratic formula to get We can factor pull out of the square root to get Here, it is easy to figure out the values of . Let , then which is note that , are integers. Examining the parity, we find that and are of the same parity. Now, we solve by factoring. We can find that and are the only positive integer values of that make a positive integer. gives and , but we can ignore the latter. gives , as well as a fraction which we can ignore.
Since and are the only two solutions and we want the sum of the third terms, our answer is . -BorealBear, minor edit by Kinglogic
To prove this, let , then which is then remembering that and are integers see if you can figure it out. -PureSwag
Solution 3
Proceed as in solution 2, until we reach Write
, it follows that for some (positive) integer k and .
Taking both sides modulo , , so .
When , we have and . When , we have and . Summing the two cases, we have .
-Ross Gao
Solution 4 (Combining Solution 1 and Solution 3)
As in Solution 1, write the three integers in the sequence as , , and .
Then the sum of the squares of the three integers is .
Setting this equal to the middle term times the common difference squared, which is ,
and solving for we get:
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, .
For to be a perfect square, must be a perfect square as well.
This means that is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be . Then:
. Now when you divide the numerator and denominator by 3, you are left with
. Because the sequence is of integers, d must also be an
integer, which means that must divide .
Taking the above equation we can solve for : .
This means that is divisible by . is automatically divisible by , so
must be divisible by . Then must be either of . Plugging back into the equation,
, so .
, so .
Finally,
-KingRavi
Solution 5
Following from previous solutions, we derive We divide both sides to get Since is an integer, must also be an integer, so we have , for some factor . We then get We then take this to modulo , getting The only possibilities for are therefore 1 and 2. We plug these into , for and , giving us the sequence , or and , for the sequence
-RYang2
Video Solution
Video Solution 1
https://www.youtube.com/watch?v=I43RH5DUa1I
Video Solution 2
https://youtu.be/M3DsERqhiDk?t=1465
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.