Problem Set #1
by Tommy2000, Nov 29, 2016, 5:29 AM
Hmm, I guess I'll record some problems I recently did/had explained to me word for word '-' I'll post sols later if I have time...
1) Show that if
is prime for 
, then 
are primitive roots.
2) Prove that for
,
![\[ \sum_{m = 1} ^ p \left( \sum_{n = 1} ^ h \left( \frac{m + n}{p} \right) \right) ^ 2 = h (p - h) \]](//latex.artofproblemsolving.com/4/f/e/4fe29ff7873efd8744184b1018d3c742090f6c81.png)
3) USAMO 2005/3
This problem isn't that fun...so I'm too lazy to post a solution.
4) WOOT 11/25 PoTD
5) USACO Dec 15 Plat #1-3
6) 2016 PUMaC Finals #1-3
7) Find all
satisfying 
and 
.
8) (2016 Turkey TST #5) Find all functions
such that 
and 
.
It's also worth noting this nice result in fractals' solution. It was something I tried to prove, but ultimately could not.
1) Show that if
is prime for 
, then 
are primitive roots.
is not a primitive root. Then 
.
.
,
is not a primitive root. Then 
.
.
,2) Prove that for
,![\[ \sum_{m = 1} ^ p \left( \sum_{n = 1} ^ h \left( \frac{m + n}{p} \right) \right) ^ 2 = h (p - h) \]](http://latex.artofproblemsolving.com/4/f/e/4fe29ff7873efd8744184b1018d3c742090f6c81.png)
.![\[ h \sum_{i = 1} ^ p \left( \frac{i}{p} \right) ^ 2 + 2 \sum_{k = 1} ^ h (h - k) S_k \]](http://latex.artofproblemsolving.com/c/4/4/c44f103ebd911d969aa45349cead8d1d1a482846.png)
In the first part, all are equal to 
unless 
,
.
to the sum.
for all 
.![\[ S_k = \sum_{i = 1} ^ p \left( \frac{i ^ 2 + ik}{p} \right) = \sum_{i = 1} ^ p \left( \frac{(i + k/2) ^ 2 - (k / 2) ^ 2}{p} \right) \]](http://latex.artofproblemsolving.com/9/4/9/9492612a13b46fe382d0fe4ed1e11a83eb7f01d3.png)
It suffices to find when 
for some 
.
(it's nonzero) to obtain an equation of the form 
where 
is nonzero. It's clear that the 
in this are bijective with the 
in the previous equation. We can write 
and 
for some 
.
or else 
.
and 
yield the same pair, we see there are 
values of 
which yield 
values of 
values of 
.
as desired.
as desired.3) USAMO 2005/3
This problem isn't that fun...so I'm too lazy to post a solution.
4) WOOT 11/25 PoTD
.
and 
so 
is cyclic, so 
lies on the perpendicular bisector of 
.
,
,
,
.
.5) USACO Dec 15 Plat #1-3
6) 2016 PUMaC Finals #1-3
be the unit circle. Then we have,
is inside the quadrilateral, we have 
iff
Hence, the two angles are equal.7) Find all
satisfying 
and 
.
is odd. Suppose 
,
and 
.
,
is either 
,
and this contradicts 
.
.
such that 
.
.
is negative. Since 
,
,
,![$f(c) \in [0, 1]$](http://latex.artofproblemsolving.com/d/6/9/d69c715e5e86ee1cd3f5364e1fd7ea53d1237f83.png)
.![$f(k) \in [\lfloor k \rfloor, \lceil k \rceil]$](http://latex.artofproblemsolving.com/0/3/2/032622cb8582feb4cfa8533d59ac1b49bf94b55e.png)
.
is increasing. Suppose 
and 
.
.
.
for integers 
are dense in reals, so we're done.8) (2016 Turkey TST #5) Find all functions
such that 
and 
.
by plugging 
in (2). Furthermore, it's clear 
for all 
and 
by (1).
for 
for prime 
.
such that 
.
.![\[ 1 + pm = qn \mid f(1) + f(pm) = 1 + f(pm) \]](http://latex.artofproblemsolving.com/5/8/9/589fe59975b6800a310e14f0e31623f7282fb017.png)
However, we know 
,
for 
with 
and 
.
.
.
,![\[ p + q ^ x \mid p ^ a + \left( q ^ x \right) ^ b \]](http://latex.artofproblemsolving.com/f/3/f/f3feb186f8518dcdcb19ea05c264ac05aef11d31.png)
Applying the euclidean algorithm, we see![\[ p + q ^ x \mid p ^ {a - b} - 1 \]](http://latex.artofproblemsolving.com/1/6/5/165b8622fabc0a1eb15f5f096ae407c306365631.png)
However, 
is arbitrary, so this is a contradiction for sufficiently large It's also worth noting this nice result in fractals' solution. It was something I tried to prove, but ultimately could not.
This post has been edited 15 times. Last edited by Tommy2000, Dec 7, 2016, 2:51 AM
Reason: Subject name
Reason: Subject name
August 2018
May 2018
Reason: RE