and 
be positive integers, and let 
be a 
grid of unit squares.
or 
rectangle. A subset 
of grid squares in 
edges of the grid squares.
.
and 
,
.
,
is composite, then all positive divisors of 
.
.
be a distribution function. Prove that for any 
,![\[
G(x) = \frac{1}{2h} \int_{x-h}^{x+h} F(t) \, dt
\]](http://latex.artofproblemsolving.com/5/5/d/55d81a7d36b0a052ddeb3e189645e43eb41f61a5.png)
of complex numbers satisfying 
and 
.
of a variate 
is defined as follows:![\[
F(x) =
\begin{cases}
A, & -\infty < x < -1, \\
B, & -1 \leq x < 0, \\
C, & 0 \leq x < 2, \\
D, & 2 \leq x < \infty.
\end{cases}
\]](http://latex.artofproblemsolving.com/0/9/0/0903946b6a44e9bfc3a1adf8dc0ad5ac013e92c1.png)
are constants. Determine the values of 
and 
of the equation 
which follows 
(b) Solve for real 
where 
![$ f:[0,1]->R$](http://latex.artofproblemsolving.com/e/c/f/ecf649bc1479dd69a79ed8389297b15557c29200.png)
bounded
Y by KevinYang2.71, aidan0626, MathRook7817, Sedro
Let 
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
Note: A function is bijective if for every integer 
, there exists exactly one integer 
such that 
.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.Note: A function
is bijective if for every integer 
, there exists exactly one integer 
such that 
.This post has been edited 1 time. Last edited by imagien_bad, Mar 20, 2025, 12:09 PM
be defined as![\[
d(x) = f(x+1) - f(x).
\]](http://latex.artofproblemsolving.com/9/6/1/9616c1dce04c5e6c8166a5c7183b29a1844b3245.png)
for some integer ![\[
\begin{aligned}
g(x+1) &= f(x+1) - (x+1)(f(x+1) - f(x)) \\
&= f(x+1) - (f(x+1) - f(x)) - x(f(x+1) - f(x)) \\
&= f(x) - x(f(x+1) - f(x)) \\
&= g(x),
\end{aligned}
\]](http://latex.artofproblemsolving.com/a/4/8/a48eab04cb0b085b0f1b20c4bfe336aeae36f5ef.png)
which implies that 
is not bijective.
.![\[
\begin{aligned}
g(x+1) &= g(x) + d(x) + c \\
&\geq g(x) + m + c \\
&\geq g(x) + 2,
\end{aligned}
\]](http://latex.artofproblemsolving.com/7/1/1/7110c01ce63fee4868242272d0b638edaa597ae2.png)
which implies that 
is bijective for all 
for some constant 
.
and 
,
for all nonnegative integers 
is bijective.
is strictly increasing.
for some integer 
.
gives![\[ g(n+1)=h(n+1)+m(n+1)=h(n)+mn=g(n), \]](http://latex.artofproblemsolving.com/7/6/7/7670d19785ce7d07f408dfb6412502173d45ab2c.png)
which contradicts our assumption that 
,
,![\[ g(x)\ge h(x)+1\ge h(0)+2, \]](http://latex.artofproblemsolving.com/f/9/5/f95080566f54d188f433e7a72b2fdd7b7af7500e.png)
while for all 
,![\[ g(x)\le h(x)\le h(0), \]](http://latex.artofproblemsolving.com/4/9/a/49a841cad1df45f76ae7ccdb892172de83c77bf7.png)
so 
,
how many mohs do yall think this is?
with 
)
.
,
to all of them, failing injectivity and thus failing bijectivity. Also just replace 
.
is much bigger than 
.
,
.
,
.
,
such that 
is not bijective for only finitely many integers 
.
can only take on a finite number of integer values.
for integers 
.
;
.
fails to be injective for this particular value of 
,
for all integers 
.
,
![\[
\mathcal D := \{f(n+1) - f(n): n\in\mathbb Z\}.
\]](http://latex.artofproblemsolving.com/a/6/b/a6b2ef3b6d0e60074aa239c7786700e0f83685b6.png)
If 
is infinite, then 
isn't injective for any 
.
.
with the property that 
implies 
is bijective.
.
gives 
which contradicts injectivity. 
; this clearly doesn't change the problem. By triangle inequality, the claim yields 
for every 
does not attain 
,
If 
where 
is not bijective.
for some constant 
.
and choose a 
-
,
so setting 
yields 
so 
where for all 
then 
in 
violates this, contradiction. BOOM.
if and only if 
and thus from here we know that 
,
then 
for all integers 
and so on we get 
for all positive integers 
is not achieved on the positive side but also notice that 
and therefore 
for all negative 
,
.
and let 
.
.
.
.
When 
so 
.
we find 
.
.
which means 
.
.
which implies that when 
.
is linear. Then we can write 
so every integer attained by 
so the problem is clear.
It suffices to show that the RHS can attain infinitely many values. For the sake of a contradiction assume otherwise.
Then connecting this point to all other points 
and taking the union yields a finite set of lines (at least 2). But by the Pigeonhole principle, some line 
on another line distinct from 
on 
Then 
so each increment is of the form 
and making 
so 
be the set of all non-zero 
where 
as 
for some 
sum of elements of B 
,
is some subset of 
(or is any element of 
and 
.
.
.
,
,
be the maximal element of 
produces 
for some 
.
,
.
for all y, and so 
