[/quote]
,
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
such that for all 
,
is a prime number, 
are natural numbers 
)
.
.
be the midpoint of 
on triangle 
.
lies on segment 
such that 
and 
on line 
such that 
.
.
there exists a unique positive integers 
such that 
is a square of a rational number.
for which![\[2^p = 2^{q-2} + q!.\]](http://latex.artofproblemsolving.com/8/8/2/88270dc4aa4a54cc8b2e525b7d9d940c300e49ac.png)
.
are positive numbers, then prove that
Prove that
Where 
,
and 
are parallel. Angle bisectors of 
and 
intersect at 
.
and 
meet again at 
.
.
and 
such that 
and 
is a prime number
Y by
I keep seeing a certain type of problem in UIL Number Sense, though I can't figure out how to do it (I aim to do it in my head in about 7-8 seconds).
The problem is x^((p+1)/2) mod p, where p is prime.
For example 11^15 mod 29
I know it technically doesn't work this way, but using fermats little theorem (on √x^(p+1)) always gives either the number itself, x, or the modular inverse, p-x.
By using the theorem i mean √x^28 mod 29 = 1, and then youre left with √x^2 mod 29 or x, but then its + or -.
I was wondering if there is a way to figure out whether its + or -, a slow or fast way if its slow maybe its possible to speed it up.
The problem is x^((p+1)/2) mod p, where p is prime.
For example 11^15 mod 29
I know it technically doesn't work this way, but using fermats little theorem (on √x^(p+1)) always gives either the number itself, x, or the modular inverse, p-x.
By using the theorem i mean √x^28 mod 29 = 1, and then youre left with √x^2 mod 29 or x, but then its + or -.
I was wondering if there is a way to figure out whether its + or -, a slow or fast way if its slow maybe its possible to speed it up.
