be circumscribed about circle 
,
be the orthocenter of 
.
at 
.
at 
.
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
.
Prove that 
When does equality hold?
,
,
and 
,
is the intersection point of the diagonals 
and 
.
and radius 
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
Y by
Positive integers 
satisfy both of the following conditions.
For a positive integer
, if 
, then 
.
There exist integers
that satisfies the equation 
and 
.
Prove that there exist integers that satisfies the equation 
, for each integer 
.
satisfy both of the following conditions.For a positive integer
, if 
, then 
.There exist integers
that satisfies the equation 
and 
.Prove that there exist integers
that satisfies the equation 
, for each integer 
.This post has been edited 4 times. Last edited by Hip1zzzil, Mar 30, 2025, 1:07 PM
Reason: Better
Reason: Better
be a solution of 
,
,
,
with 
gives all 
,
and 
are square free and coprime.
so that 
.![\[ (\zeta t + p)^2 + (\omega t +q)^2 -a(\alpha t +r)^2 -b(\beta t +s)^2 = 2(\zeta p + \omega q -a\alpha r -b \beta s)t + p^2 +q^2 -ar^2 -bs^2 \]](http://latex.artofproblemsolving.com/9/c/c/9ccadb23a0effe0bce5929a9ee9297a88f46e93a.png)
From Bezout, we have 
such that![\[ \zeta p + \omega q -a\alpha r -b \beta s =1 \]](http://latex.artofproblemsolving.com/e/d/5/ed551a451a3e968217b90a7aee89058b1d687ff3.png)
so we have all ever, or all odd 
,
,
are all odd, so for that case I used ![\[ \zeta p + \omega q -a\alpha r -b \beta s =2 \]](http://latex.artofproblemsolving.com/0/c/e/0ce66f661dbb7eb85a13ba230006a2c72b67cdde.png)
and eventually I got all odds, 
s
s
is not squarefree then 
is a positive integer...
