Difference between revisions of "2022 USAJMO Problems/Problem 1"
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− | Note: This shouldn't work since we see that m = 12 is a solution. Let the initials for both series by 1, then let the ratio be 7 and the common difference to be 6. We see multiplying by 7 mod 12 that the geometric sequence is alternating from 1 to 7 to 1 to 7 and so on, which is the same as adding 6. Therefore this solution is wrong. My counter-conjecture is that all non square-free m (4, 8, 9, 16, 18, 25...) should all work, but I don't have a proof. However, if you edit the one above, you can see | + | Note: This shouldn't work since we see that m = 12 is a solution. Let the initials for both series by 1, then let the ratio be 7 and the common difference to be 6. We see multiplying by 7 mod 12 that the geometric sequence is alternating from 1 to 7 to 1 to 7 and so on, which is the same as adding 6. Therefore this solution is wrong. My counter-conjecture is that all non square-free m (4, 8, 9, 16, 18, 25...) should all work, but I don't have a proof. However, if you edit the one above, you can see non square-free m will work. In order to construct a ratio, we could us (4) and find a square multiple of m, take the square root and add 1 to get the ratio. |
==See Also== | ==See Also== | ||
{{USAJMO newbox|year=2022|before=First Question|num-a=2}} | {{USAJMO newbox|year=2022|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:59, 5 April 2024
Problem
For which positive integers does there exist an infinite arithmetic sequence of integers
and an infinite geometric sequence of integers
satisfying the following properties?
is divisible by
for all integers
;
is not divisible by
.
Solution 1
We claim that satisfies the given conditions if and only if
is a perfect square.
To begin, we let the common difference of be
and the common ratio of
be
. Then, rewriting the conditions modulo
gives:
Condition holds if no consecutive terms in
are equivalent modulo
, which is the same thing as never having consecutive, equal, terms, in
. By Condition
, this is also the same as never having equal, consecutive, terms in
:
Also, Condition holds if
Restating, , and the conditions
and
hold if and only if
is a perfect square.
[will finish that step here]
Note: This shouldn't work since we see that m = 12 is a solution. Let the initials for both series by 1, then let the ratio be 7 and the common difference to be 6. We see multiplying by 7 mod 12 that the geometric sequence is alternating from 1 to 7 to 1 to 7 and so on, which is the same as adding 6. Therefore this solution is wrong. My counter-conjecture is that all non square-free m (4, 8, 9, 16, 18, 25...) should all work, but I don't have a proof. However, if you edit the one above, you can see non square-free m will work. In order to construct a ratio, we could us (4) and find a square multiple of m, take the square root and add 1 to get the ratio.
See Also
2022 USAJMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.