Difference between revisions of "2022 USAJMO Problems/Problem 1"

Latest revision as of 04:52, 10 March 2025

Problem

For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1,a_2,\cdots$ and an infinite geometric sequence of integers $g_1,g_2,\cdots$ satisfying the following properties?

$\bullet$ $a_n-g_n$ is divisible by $m$ for all integers $n>1$ ;

$\bullet$ $a_2-a_1$ is not divisible by $m$ .

Solution

Let the arithmetic sequence be $\{ a, a+d, a+2d, \dots \}$ and the geometric sequence to be $\{ g, gr, gr^2, \dots \}$ . Rewriting the problem based on our new terminology, we want to find all positive integers $m$ such that there exist integers $a,d,r$ with $m \nmid d$ and $m|a+(n-1)d-gr^{n-1}$ for all integers $n>1$ .

Note that $m | a+nd-gr^n \; (1),$ $m | a+(n+1)d-gr^{n+1} \; (2),$ $m | a+(n+2)d-gr^{n+2} \; (3),$

for all integers $n\ge 1$ . From (1) and (2), we have $m | d-gr^{n+1}+gr^n$ and from (2) and (3), we have $m | d-gr^{n+2}+gr^{n+1}$ . Reinterpreting both equations,

$m | gr^{n+1}-gr^n-d \; (4),$ $m | gr^{n+2}-gr^{n+1}-d \; (5),$

for all integers $n\ge 1$ . Thus, $m | gr^k - 2gr^{k+1} + gr^{k+2} = gr^k(r-1)^2 \; (6)$ . Note that if $m|g,r$ , then $m|gr^{n+1}-gr^n$ , which, plugged into (4), yields $m|d$ , which is invalid. Also, note that (4) $+$ (5) gives

$m | gr(r-1)(r+1) - 2d \; (7),$

so if $r \equiv \pm 1 \pmod m$ or $gr \equiv 0 \pmod m$ , then $m|d$ , which is also invalid. Thus, according to (6), $m|g(r-1)^2$ , with $m \nmid g,r$ . Also from (7) is that $m \nmid g(r-1)$ .

Finally, we can conclude that the only $m$ that will work are numbers in the form of $xy^2$ , other than $1$ , for integers $x,y$ ( $x$ and $y$ can be equal), ie. $4,8,9,12,16,18,20,24,25,\dots$ .

~sml1809

Solution 1

$m$ satisfies the conditions precisely when $m$ is not squarefree.

Consider three consecutive terms of the arithmetic sequence modulo $m$ : $x - d$ , $x$ , and $x + d$ . To satisfy the given conditions, the arithmetic and geometric sequences must match modulo $m$ . Thus, $(x - d)(x + d) \equiv x^2 \pmod{m}$ which simplifies to $x^2 - d^2 \equiv x^2 \pmod{m} \implies d^2 \equiv 0 \pmod{m}.$

If $m$ is squarefree, the congruence $d^2 \equiv 0 \pmod{m}$ implies $m \mid d$ , so the conditions are not satisfied. Thus, no solution exists for squarefree $m$ , and it remains for us to construct a solution for all other values of $m$ that are not squarefree.

Suppose $m$ is divisible by some prime square $p^2$ . Consider the arithmetic sequence $1,\quad 1 + \frac{m}{p},\quad 1 + 2\frac{m}{p},\quad \dots$ and the geometric sequence $1,\quad 1 + \frac{m}{p},\quad \left(1 + \frac{m}{p}\right)^2,\quad \dots$ For the geometric sequence, the first two terms in the binomial expansion of each entry match the corresponding terms in the arithmetic sequence and subsequent terms are 0 mod m because $\frac{m}{p}$ raised to a power greater than 1 contains enough $p$ 's in its factorization to ensure that it is a multiple of $m$ . Therefore the construction works and as originally stated, solutions exist precisely when $m$ is not squarefree.

@@ Line 32: / Line 32: @@
 ==Solution 1==
-(This claim below is false, numbers like 12 work, or anything indicated in the previous solution)
+<math>m</math> satisfies the conditions precisely when <math>m</math> is not squarefree.
-We claim that <math>m</math> satisfies the given conditions if and only if <math>m</math> is a perfect square.
-To begin, we let the common difference of <math>\{a_n\}</math> be <math>d</math> and the common ratio of <math>\{g_n\}</math> be <math>r</math>. Then, rewriting the conditions modulo <math>m</math> gives:
+Consider three consecutive terms of the arithmetic sequence modulo <math>m</math>: <math>x - d</math>, <math>x</math>, and <math>x + d</math>. To satisfy the given conditions, the arithmetic and geometric sequences must match modulo <math>m</math>. Thus,
-<cmath>a_2-a_1=d\not\equiv 0\pmod{m}\text{         (1)}</cmath>
+<cmath>(x - d)(x + d) \equiv x^2 \pmod{m}</cmath>
-<cmath>a_n\equiv g_n\pmod{m}\text{             (2)}</cmath>
+which simplifies to
+<cmath>x^2 - d^2 \equiv x^2 \pmod{m} \implies d^2 \equiv 0 \pmod{m}.</cmath>
-Condition <math>(1)</math> holds if no consecutive terms in <math>a_i</math> are equivalent modulo <math>m</math>, which is the same thing as never having consecutive, equal, terms, in <math>a_i\pmod{m}</math>. By Condition <math>(2)</math>, this is also the same as never having equal, consecutive, terms in <math>g_i\pmod{m}</math>:
+If <math>m</math> is squarefree, the congruence <math>d^2 \equiv 0 \pmod{m}</math> implies <math>m \mid d</math>, so the conditions are not satisfied. Thus, no solution exists for squarefree <math>m</math>, and it remains for us to construct a solution for all other values of <math>m</math> that are not squarefree.
-<cmath>(1)\iff g_l\not\equiv g_{l-1}\pmod{m}\text{ for any integer }l>1</cmath>
+Suppose <math>m</math> is divisible by some prime square <math>p^2</math>. Consider the arithmetic sequence
-<cmath>\iff g_{l-1}(r-1)\not\equiv 0\pmod{m}.\text{        (3)}</cmath>
+<cmath>1,\quad 1 + \frac{m}{p},\quad 1 + 2\frac{m}{p},\quad \dots</cmath>
+and the geometric sequence
+<cmath>1,\quad 1 + \frac{m}{p},\quad \left(1 + \frac{m}{p}\right)^2,\quad \dots</cmath>
-Also, Condition <math>(2)</math> holds if
+For the geometric sequence, the first two terms in the binomial expansion of each entry match the corresponding terms in the arithmetic sequence and subsequent terms are 0 mod m because <math>\frac{m}{p}</math> raised to a power greater than 1 contains enough <math>p</math>'s in its factorization to ensure that it is a multiple of <math>m</math>. Therefore the construction works and as originally stated, solutions exist precisely when <math>m</math> is not squarefree.
-<cmath>g_{l+1}-g_l\equiv g_l-g_{l-1}\pmod{m}</cmath>
-<cmath>g_{l-1}(r-1)^2\equiv0\pmod{m}\text{        (4)}.</cmath>
-Restating, <math>(1),(2)\quad \textrm{if} \quad(3),(4)</math>, and the conditions <math>g_{l-1}(r-1)\not\equiv 0\pmod{m}</math> and <math>g_{l-1}(r-1)^2\equiv0\pmod{m}</math> hold if and only if <math>m</math> 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. Let <math>m = at^2</math> then <math>at + 1 \not\equiv 1 \pmod{at^2}</math> or <math>at</math> is not divisble by <math>at^2</math>. If <math>t = 1</math>, this is false and this is not possible. But if it isn't, if <math>m</math> isn't square free, then it should work.
-Note: This counter-conjecture is correct. To prove it, it suffices to show that if m is square-free, then (3) and (4) contradict each other. Indeed, if m is square-free, then each prime dividing m only has a power of 1 in the prime factorization, so given (4) that m|<math>x\cdot (r-1)^2</math>, m has at most one of each prime factor of <math>x \cdot (r-1)^2</math>, but then m has at most one of each prime factor of <math>x \cdot (r-1)</math>, so m divides <math>x \cdot (r-1)</math>, contradicting (3).
 ==See Also==
 {{USAJMO newbox|year=2022|before=First Question|num-a=2}}
 {{MAA Notice}}

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Difference between revisions of "2022 USAJMO Problems/Problem 1"

Latest revision as of 04:52, 10 March 2025

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Problem

Solution

Solution 1

See Also