AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
. Consider a function 
. A subset 
of 
is said to be invariant if for all 
we have 
. The empty set and are also considered as invariant subsets. By 
we define the number of invariant subsets of for the function 
. such that 
.
there exists a function such that 
.
and 
for 
. Let 
be a positive integer. Suppose that for every positive integer 
there exists a positive integer 
such that 
. Must be a Fibonacci number?
satisfying following conditions.
for all 
.
with 
, equality 
holds.
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
be the circle 
. The line 
and the circle of radius centered at 
cross at 
and 
, respectively. Prove that , the circle on diameter 
and circle 
are concurrent.
of functions from the set of real numbers to itself that satisfy ![\[g(f(x+y)) = f(x) + (2x + y)g(y)\]](http://latex.artofproblemsolving.com/0/a/3/0a3bc25299a0bcbd3f3602a0e6881d68fce750c5.png)
for all real numbers 
and 
.
such that for all 
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call to be a grandson of if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
of primes (greater than 5) such that : 
of positive integers such that:
has incentre 
and incircle touching 
at . Let 
be the antipode of in the circumcircle of . Point 
is chosen on the internal angle bisector of 
such that 
. Let 
be the midpoint of arc 
, and let 
be the midpoint of 
. Prove that 
is a power series for which each coefficient 
is 
or 
. Show that if 
, then 
must be irrational.
equal to . From there, that shows that if the sum converges with denominators of powers of 
to a quantity with a non-common denominator of 
, it will be irrational for an initial denominator of ...that's a rough sketch at least.
is rational, then 
is a rational with odd denominator.![\[f(1/2) = c_0.c_1c_2c_3 \dots_{(2)}\]](http://latex.artofproblemsolving.com/7/8/5/78526b3a87f10eb2afd5734f1e4ce4b395b7e2ec.png)
is rational, then 
is eventually periodic: there exist positive integers 
such that 
whenever 
. Then
a sum of rationals with odd denominators; consequently is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)
in the denominator, I was probably on the right track?
has to be periodic for to be rational?
. Does anyone have results on how setting 
for rational x, y determines the rationality/irrationality of other f(x)=y?
has to be periodic for to be rational? is rational then the sequence is eventually periodic?" I had most of the proof of this in the exam but I skipped the final computation (basically some Euler's function involved...ew...)
be rational number. Multiplying and dividing by powers of does not change whether the sequence is eventually periodic or not, so assume 
with 
and 
positive integers and gcd(, , 2) = . By Fermat's Little Theorem there exists a positive integer 
with 
divisible by . Then 
is an integer, so the binary expansion of 
is periodic with period .
. The contrapositive/base-2 expansion method is the way to go. My proof is essentially identical to CantonMathGuy's. formed a terminating or eventually repeating sequence, and showed that the equation was impossible (due to denominator arguments). Then I said by contradiction, f(1/2) is irrational
formed a terminating or eventually repeating sequence, and showed that the equation was impossible (due to denominator arguments
rational.
since all fractions have denominators with no powers of while 
. Would that be considered correct? Just in case, I derived the basic theory of valuations like 
. But I guess this might actually be an overkill.
is rational, then is a rational with odd denominator.is rational, then is eventually periodic: there exist positive integers such that whenever . Thena sum of rationals with odd denominators; consequently is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)
in any case? It seems to me that 
is always a sum of rationals with odd denominator anyways.
, 
, 
, 
, and so on; 
is if the remainder is greater than 
, and zero otherwise.
. (Any farther, and I'm afraid of floating-point errors creeping in)
is rational, then is a rational with odd denominator.is rational, then is eventually periodic: there exist positive integers such that whenever . Thena sum of rationals with odd denominators; consequently is a rational with odd denominator. (We may rearrange the terms in the sum because it contains only nonnegative terms and converges; hence converges absolutely.)
, we have that 
because the denominator is always a power of . Now, this means that there are an infinite amount of such that 
. Now, we assume the contrapositive: 
is rational. Writing in binary decimal, this means that there repeats, meaning after a number then there exists some such that for all 
, 
. This means that, letting 
gets that our sum is 
and since that expression's denominators are all off and the sum is finite, then 
of that expression is nonnegative. However, 
which reaches a contradiction.
is rational. Then, interpret the value of the series as the value of a binary decimal. This decimal represents a rational number if and only if it is eventually periodic, by definition. Hence, the series must be eventually periodic.
is periodic for all 
, in particular with 
. Furthermore, let the period of repetition be 
. Then, our sequence can be split into two parts: the repeating part and the nonrepeating part. Note that by definition, the sum of the repeating part is 
The 
of the inner expression is at least 
by definition, so the of the entire expression must also be . of the nonrepeating part is nonnegative; thus, the sum of the series must also have a nonnegative , contradiction. Thus, is necessarily irrational.
cannot be eventually periodic. FTSOC assume otherwise. Then, let and 
be such that for 
, we have 
We then have 
Let's rewrite this: 
Since 
is odd, and all the other denominators on the left side are 3, there is no way to get a even number in the denominator for the LHS, contradiction. Therefore, the sequence cannot be eventually periodic, so is irrational.
is rational, we have![\[f(1/2)=c_0.c_1c_2c_3\dots_{2}\]](http://latex.artofproblemsolving.com/0/0/9/009d23e747e05abb497832fc1a9b6c7cee1e79b3.png)
, we have 
. Then,
is periodic. Thus if is the period of , ![\[ \frac{3}{2} \cdot \left( \frac{3^k-2^k}{3^k} \right) \]](http://latex.artofproblemsolving.com/6/f/e/6fea59072f387b61fc0e622c15a2cb868efbac44.png)
is equal to a finite decimal in base 
. But this is impossible since the denominator in reduced form is even, as desired.
it has either a terminating or eventually repeating binary expansion (!!). This is in fact the main difficulty of the problem.
we have 
. However now as 
and 
. Now as,
as long as 
so clearly this case fails. is eventually repeating. Assume it has an initial string of 
characters, and then repeats with period . Then we need,
for some integer constants 
and not divisible by . From sum of geometric series we find,
Then we must have,
However now examining the we have a contradiction as the right hand side can never have an even denominator.
is explicit. In this case, in its binary expansion (resulting from the given order), we would have a periodicity of coefficients from a point onwards. So there are polynomials with coefficients 
in 
such that![\[
f(x) = P(x) + \left(1 + x^N + x^{2N} + \ldots\right)Q(x) = P(x) + \frac{Q(x)}{1 - x^N}
\]](http://latex.artofproblemsolving.com/e/f/9/ef956fc894bb9fdd2faa5d694c7af9196726dba6.png)
, the left member is equal (given) to 
. But the right, making fractions homonymous separately in and adding, is of the form![\[
\frac{A}{3^M} + \frac{B}{(3^N - 2^N)3^K}
\]](http://latex.artofproblemsolving.com/f/6/5/f65806238d58ba3778ae9f6155ba11b20628e623.png)
. Multiplying now by the least common multiple 
of the two denominators on the right, which is an odd number, follows 
an integer. Out of place, and so on.
is rational, then its decimals must eventually be periodic in all bases. Considering it in base , for some starting point 
, some period , some ![$i \in [0, k-1]$](http://latex.artofproblemsolving.com/d/3/0/d309acd1d73169f02b82f58c14ed09aac2ec1042.png)
, 
for all nonnegative integers 
. Thus, we know that must be eventually periodic, which transforms 
into a finite sum.
. For all 
we have 
. That means
of each term in the sequence are 
at max (if they even contribute any to the sum at all). That means the of the sequence can never drop below by the strong triangle inequality. Thus, 
, a contradiction.
is rational. Since 
is just the binary representation of , it's well-known that the sequence 
is eventually periodic. Suppose the sequence begins to repeat at 
with period (it's possible that 
). Then we have ![\[f\left(\frac23\right) = \sum_{n = 0}^{k - 1} c_n\left(\frac23\right)^n + \frac{3^r}{3^r - 2^r} \sum_{n = k}^{k + r - 1} c_n\left(\frac23\right)^n,\]](http://latex.artofproblemsolving.com/9/7/1/971fdf713e83ad2f6c0ada6fd7689d611a9c53ce.png)
so since 
is clearly , we have ![\[\nu_2(f(2/3)) \geq \min\left(\nu_2\left(\sum_{n = 0}^{k - 1} c_n\left(\frac23\right)^n\right), \nu_2\left(\sum_{n = k}^{k + r - 1} c_n\left(\frac23\right)^n\right)\right) \geq 0,\]](http://latex.artofproblemsolving.com/d/b/c/dbcec0458fa6edb0a2fac946f59fee91b7626b6c.png)
a contradiction. 
. This is equivalent to the binary decmal 
being rational. Therefore this means that 
must be eventually periodic. has period length and a new cycle starts at 
. By the geometric series formula we require
Observe that the LHS is the sum of finitely many fractions with odd denominators. Therefore the LHS can be written as a fraction with an odd denominator. However the RHS has an even denominator. Contradiction.
is rational. This implies the must be eventually periodic. It can't have a finite number of s clearly, since 
would have a denominator that's a power of , so the repeating part must be nonzero. Let 
be the beginning part that doesn't repeat. Then, the part that does repeat can be written as 
. Now, it's clear that the denominator of is odd, so it can't possibly simplify to 
.
representation of 
it is clear that is rational iff the sequence 
is eventually periodic.
be the smallest integers such that 
for all 
.
where 
.
. But 
is always odd, contradiction.
be such that![\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]](http://latex.artofproblemsolving.com/1/5/e/15e5bf0cc14794e7028dbfe4a8754f6803e850ea.png)
and 
.
.
is rational, so that 
is eventually periodic. Then, 
is the sum of finitely many 
'![\[\left(\frac{2}{3}\right)^m+\left(\frac{2}{3}\right)^{m+n}+\dots=\frac{\left(\frac{2}{3}\right)^m}{1-\left(\frac{2}{3}\right)^n}.\]](http://latex.artofproblemsolving.com/5/3/0/53047645ef824b8a499605553a6bee2cca542891.png)
All these expressions (finite in number) have nonnegative 
.
for some 
.