AoPS Academy
such that 
. Show that the equation 
has no integer solution.
denotes a polynomial of degree 
such that 
for 
, determine 
.
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. 
is the orthocenter. 
is the midpoint of segment 
. Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from 
and 
intersect at 
. Prove that 
are colinear
be a linear transformation on linear space 
satisfying:
but 
, and 
is the eigensubspace of eigenvalue 
. Prove that can be decomposed to 
-cyclic subspace's direct sum. be a square matrix of odd order (at least 
) whose entries are odd integers. Prove that if is invertible, then it is not possible for all the minors of the entries of a row to have equal absolute values.
be then 
-th fibbiance number. As gets bigger and bigger, we have,
my question is dose,
My reservations about this is that 
and 
so 
. So, if the limit holds, does that mean that if 
is a set and 
is a set, for each 
that 
we can have, for 
we can have,
for some ?
+1 w
such that 
for all 
order, and if the constituents of its first, second, ..., rows are the first figurate numbers of the first, second, ..., orders respectively, show that it's value is 
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
,
with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
and 
.
be the assertion that 
.
,
s.t. 
.
we get 
,
.
yields 
.
and using 
,
.
is even we can use again 
.
,
.
.
we get 
,
,
,
.
,
is odd, by 
,
;
,
,
;
,
,
.
,
,
,
.
is the set of all integer multiples of some nonnegative integer 
,
,
.
,
.
,
is a surjective map from integers to integers. Substituting, we have that 
.
,
can be any integer. Setting 
,
.
for some 
,
,
,
and 
we have that 
,
,
,
.
,
or 
,
and 
.![\[f(x-f(y))=f(f(x))-f(y)-1.\]](http://latex.artofproblemsolving.com/7/0/9/70926ba51d8508d458d1b5ef9b15358f0f4040ea.png)
![\[ f(x-f(f(x)))=-1, \]](http://latex.artofproblemsolving.com/4/4/0/440aefc6e12e5f2791d60e1ad62babe7eb3ad67b.png)
so 
must be in the range of 
.![\[ f(x+1)=f(f(x)). \quad (\star). \]](http://latex.artofproblemsolving.com/3/e/7/3e732fd1a733e0df0364d4862fe4c93a1a98e4c0.png)
Now from 
we find that 
implying that 
is constant for all 
.
,
must be of the form 
for some constants 
.
,![\[ k(x+1)+c = k(kx+c)+c, \]](http://latex.artofproblemsolving.com/2/7/4/27416211ae45c11638424f0c2950fb707e26bdee.png)
so ![\[ k(x+1)=k(kx+c). \]](http://latex.artofproblemsolving.com/3/1/7/31772e564c707726f1b3c998888f10175c58b2c7.png)
If 
,![\[ kx+c=x+1, \]](http://latex.artofproblemsolving.com/b/b/f/bbf542d003ec3440fbe46298120b505ef5199b66.png)
meaning that ![\[ c=c-c-1=-1,\]](http://latex.artofproblemsolving.com/f/4/a/f4ab4c35611a39c019fc5188d1823ae631008f10.png)
so 
are indeed the only solutions to 
for all 
,
for all 
,
,
in 
,
So by (2), choose 
gives 
,
Applying this successively we obtain 
With this 
.
,
we get 
so take 
,
.
,
Since 
for all 
.
,
for all 
for all 
.
in terms of 
.
,
,
,
.
We split into 
cases.
.
,
for some 
,
,
In particular, 
,
,
,
,
takes more than 
value, which is impossible since 
,
for all 
,
,
.
,
for all 
,
,
![\[f(x+1)=f(f(x))\]](http://latex.artofproblemsolving.com/7/5/9/759f6e67ec4c8f32633a751392b492e56f84b418.png)
![\[f(-1)=f(f(f(x)-1))-f(x)-1=f(f(x))-f(x)-1=f(x+1)-f(x)-1\longrightarrow f(x+1)-f(x)=f(-1)+1\]](http://latex.artofproblemsolving.com/7/3/5/735a45eee465c8c3101bd853bd7b52562562ac04.png)
Thus 
which also works.
such that 
.
shows that 
.
gives 
,
and solve. 
,
and 
.
or 
.
.
we have 
.
such that 
for 
distinct. We claim this gives 
and 
gives 
,
.
gives 
.
is in the range of 
.
are both in the range of 
and 
.
gives our contradiction, as 
yields 
yields 
,
yields 
.
and going in the negative and positive direction.
yields, using 
,
which completes the inductive step. In the negative direction, setting 
yields, again using 
which completes the inductive step, proving Lemma 1.
or 
.
,
.
for all integers 
if 
.
.
is nonnegative, both 
and 
are positive and strictly increasing as 
,
and 
.
yields 
via 
yields via 
We know 
from 
and 
.
and, with 
.
yields that for all integers 
.
and, with 
.
.
yields that for all positive integers 
for any 
.
.
,
are positive integers and 
,
.
such that 
.
and 
for all 
is constant.
and 
gives ![\[f(f(b)) = f(f(f(a)+f(b))) - f(a) - 1.\]](http://latex.artofproblemsolving.com/a/6/3/a63d2d5ae4fdf39c3f0233b21d5e6707c5c58c0c.png)
Taking 
gives ![\[f(f(a)) = f(f(f(a)+f(b))) - f(b) - 1.\]](http://latex.artofproblemsolving.com/c/8/d/c8ddf46ee753c8ce268b304c22e8d83aaba30902.png)
Comparing the two equations gives 
,
for some constant ![\[f(x-f(y)) = f(x) - f(y) + (k-1)\]](http://latex.artofproblemsolving.com/0/4/a/04aef906b57083d0aec19b9818947fa990520385.png)
for all 
.
,![\[f(x+1) = f(x) + k\]](http://latex.artofproblemsolving.com/f/9/7/f97a4006a36721681c7d4ce75d8bde2b307ac094.png)
for all 
for some constants ![\[k(x-ky-c)+c = k(kx+c) + c - ky - c - 1\]](http://latex.artofproblemsolving.com/d/3/0/d30e23d9461f8037178685dd7701cf2770413438.png)
or ![\[kx - k^2y - ck + c = k^2x + ck - ky - 1.\]](http://latex.artofproblemsolving.com/b/e/9/be996fdee2d03f3f7a96b2b48cac607f84e93382.png)
Comparing 
,
.
,
,
;
.
.
.
,![\[f(x-f(f(x))) = -1,\]](http://latex.artofproblemsolving.com/d/1/f/d1f22cb4a23811569411a3f97761c44b36cbe53b.png)
.
:![\[f(x+1) = f(f(x)).\]](http://latex.artofproblemsolving.com/1/c/a/1cab33aa5c0233e3c279fd7b8ffda1aa1c6faa9c.png)
.
yields![\[f(k)+1 = f(x+k+2)-f(x).\]](http://latex.artofproblemsolving.com/e/b/7/eb73f596a083386b530923214e106e8ca6ae4d0c.png)
shows that each consecutive difference is constant. Hence, 
and solving gives the solution set.
implies 
.
,
so we get 
.
.
This implies that either 
where c is a constant value and 
.
.
is a constant value, and then we can write 
.
and then we can vary the value of 
where 
.
is also a solution which can be checked. Q.E.D
)
,
.
.
This proves the claim. 
that is 
,
.
.
![\[f(f(x))=f(x+1)\]](http://latex.artofproblemsolving.com/e/2/1/e214ad7b7679eb1d9d0d9a62d826e27453d17356.png)
Now substitute this back into the given FE and let 
such that 
,
from 
yields 
.
.
,
in 
.
and 
.
gives that 
,
we get 
,
.
,
for all 
.
gives ![\[-1=f(y+kc-1-f(y)), \ \forall y\]](http://latex.artofproblemsolving.com/6/a/1/6a138e6b4c19a9cf45dde9a3f2cdac3553edf470.png)
From now, the idea is to find a smaller period of 
,
with 
which gives a smaller period.
is constant modulo 
.
.
(Here 
)
is at least 
since we have 
positive and 
,
,
,
.
,
otherwise 
and 
,
shows 
,
.![\[-1=f(f(y) - f(f(f(y)))) = f(-1-f(0))\]](http://latex.artofproblemsolving.com/d/6/3/d638dd54f8ca7a08f825abb7af602fc1bce72ebf.png)
i.e. for 
,
.
in the original to get 
.![\[f(y+2) - f(y) = f(0)+1.\]](http://latex.artofproblemsolving.com/f/8/3/f8370d09d42e59db2f01437e239f4c296fd1bc04.png)
In other words, 
and 
for all integers 
for now; it follows that there are at most two values 
yields 
.
,
can take on infinitely many values, hence we must have 
.
.
or 
by setting 
,
in the original,![\[2(k-\ell) - r + 2 = f(2k+1-2\ell - r) = 2k+2r-1 - 2\ell - r - 1 = 2(k-\ell) + r-2.\]](http://latex.artofproblemsolving.com/1/6/f/16fc68f889a033133baa573e74920455221dccb7.png)
It follows 
.
for all 
yields 
,
,![\[f(1) = f(x-f(1)) = f(f(x)) - f(1) - 1 = -f(1) - 2\]](http://latex.artofproblemsolving.com/b/0/6/b066c6fe61b3ee710d30df4c379c844ae9c836ed.png)
which yields 
both of which clearly work.
be the assertion.
By subtracting, we get 
Taking 
,
Hence, 
is a linear function since its finite difference is constant. We write 
or 
plug this in to get 
Thus, 
If 
and if 
gives 
Letting 
we see that 
Then 
This implies that 
We now consider two cases.
The above boxed equation immediately gives 
Let 
for integers 
gives 
yielding our first solution 
gives 
a contradiction.
Then the first boxed equation implies that if 
then 
By the second boxed equation, we know 
is odd, as is 
Hence 
immediately giving 
Hence 
for even 
for odd 
Hence either 
or 
Both cases give 
yielding our second solution 
for all 
to get that there exists 
such that 
to get that 
which indeed works.
for some 
Note that we can get 
and so thus, 
for some 
to get that 
is linear and 
is linear. By 
we can also get that 
for two constants(not necessarily the same). Finally, we do parity casework on 
then taking 
tells us 
contradiction.
then taking 
tells us 
contradiction.
taking 
tells us 
which finishes.
![$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}}
P(f(x),x): f(0) = \underline{f(f(f(x))) - f(x) - 1 \quad (\RomanNumeral{1})} \\
P(x, f(x)): \underline{f(x - f(f(x))) = -1 \quad (\RomanNumeral{2})} \\
P(x, x - f(f(x))): f(x - f(x - f(f(x)))) = f(f(x)) - f(x - f(f(x))) - 1 \xrightarrow{\RomanNumeral{2}} \underline{f(x+1) = f(f(x)) \quad (\RomanNumeral{3})}$](http://latex.artofproblemsolving.com/c/0/e/c0e09d966290c0fad24ddc8ddf6414f88d31d96c.png)
![$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{3})$](http://latex.artofproblemsolving.com/1/a/b/1ab108061d3d10b96f34a6445db72a336cc13247.png)
we have that 
,![$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{1})$](http://latex.artofproblemsolving.com/b/b/0/bb0e47f0a77ef04a920e561d9b4a88bacc5b17ba.png)
we have: ![$ \underline{f(0) + 1 = f(x+2) - f(x) \newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} \quad (\RomanNumeral{4})}$](http://latex.artofproblemsolving.com/1/c/7/1c715c470e516b4ad9538ff107cee63cd4e97639.png)
.
is linear. So 
,
,
.
is also linear, so 
,
is odd and 
.
.
and we can cross it out: 
,![$f(2x+1) = ax+c \xrightarrow{\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} \RomanNumeral{3}} f(f(2x)) = ax + c \rightarrow f(ax + b) = ax + c \xrightarrow{x := 0} f(b) = c$](http://latex.artofproblemsolving.com/f/7/6/f761557a3440a7266a8088a363f195d54414e453.png)
so 
or 
.
and 
.
and 
is even so 
.
is the solution in this case, which indeed works.
.
.![$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{2})$](http://latex.artofproblemsolving.com/b/e/8/be8e7b4f560e33715c589527faa547f954ce1aeb.png)
we know ![$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} f(2x - f(f(2x)) = -1 \xrightarrow{\RomanNumeral{3}} f(2x - f(2x + 1)) = -1 \rightarrow f(2x - 2x - c) = -1 \rightarrow f(-c) = -1$](http://latex.artofproblemsolving.com/c/3/f/c3f6ebde7b768927d600d5dee37aa1f188e36342.png)
.
,
,
,
so 
but we supposed c is odd, contradiction.
,
in this case, which is also a solution.
.
to get![\[f(x - f^2(x)) = -1\]](http://latex.artofproblemsolving.com/9/b/a/9ba2e37465b0770d4792ffc73c3e7083158ade84.png)
Afterwards let 
,
![\[f(x+1) = f^2(x)\]](http://latex.artofproblemsolving.com/4/e/e/4eea98a4ae40573a3c52bdf0ef89b4cd8b5ca8ae.png)
Notice that the equation can be transformed into![\[f(x-f(y)) = f(x+1) - f(y) - 1\]](http://latex.artofproblemsolving.com/f/f/1/ff118d864b89b1169ae7f45346808330b1578040.png)
Let 
(otherwise 
,
,![\[f(p+1) = f(k+p+2) -k-1 \Longrightarrow 2k+1 = f(k+p+2)\]](http://latex.artofproblemsolving.com/0/2/2/0228164df9d0c06c8016e3760cbd167a78df7639.png)
Since 
,
and 
is also in the image of 
,
is also in the image and so on. Therefore the set containing all the integers which are in the image of 
be a number such that 
for some 
.
we have![\[f(x-k) = f(x+1)-k-1\]](http://latex.artofproblemsolving.com/9/d/c/9dcf3cfa71d5151d463518880af47bc6b9e24045.png)
we already know that if 
,
to get![\[f(x-2k-1) = f(x+1) - 2k - 2\]](http://latex.artofproblemsolving.com/4/d/2/4d2d873e1f5faefdce3261479519962899e8b514.png)
Thus we must have![\[f(x-2k-1) + k + 1= f(x-k) \Longrightarrow f(x) = f(x-k-1) +k+1\]](http://latex.artofproblemsolving.com/a/f/8/af815cd888fc2bc35cdecbc08abe16761495c1e3.png)
for some 
,
.
and 
and 
,![\[f(a+1) = f^2(a) = f^2(b) = f(b+1)\]](http://latex.artofproblemsolving.com/8/f/7/8f7fcf4aa848e6dfdbc9e9d2a384d818255ab64c.png)
similarly![\[f(a+2) = f^2(a+1) = f^2(b+1) = f(b+2) \Longrightarrow f(a) = f(b) = f(2b-a)\]](http://latex.artofproblemsolving.com/4/1/0/410ae8f33fd71bb72dddf9e0b915cde402a5e308.png)
By induction we have![\[f(a) = f(xb-(x-1)a)\]](http://latex.artofproblemsolving.com/7/f/e/7fecabc37ae0f18fb495fb060493de7c778f37ee.png)
However substituting 
, + a)\]](http://latex.artofproblemsolving.com/8/3/d/83df5e7fbf3f9f486e0d207c9da757214488c232.png)
But since 
, + a) = f(a) + (k+1)(b-a)\]](http://latex.artofproblemsolving.com/9/0/8/908b8d633a2c2b576c104c09c0b1d3990dddb58c.png)
This is a clear contradiction.
,
be the main equation plugging 
and 
.
,
,
.
,
,