Difference between revisions of "User:Ddk001"
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Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>. | Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>. | ||
Then, <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)</math>, so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1 \implies (p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>. | Then, <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)</math>, so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1 \implies (p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>. | ||
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+ | ==Problem 2== |
Revision as of 20:05, 1 January 2024
Contents
Problems
See if you can solve these:
1. There is one and only one perfect square in the form
where and
are prime. Find that perfect square.
2. Suppose there is complex values and
that satisfy
Find .
3. Suppose
Find the remainder when is divided by 1000.
4. Suppose is a
-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are
roots, say
. Suppose all integers
ranging from
to
satisfies
. Also, suppose that
for an integer . If
is the minimum possible value of
.
Find the number of factors of the prime in
.
5. (Much harder) is an isosceles triangle where
. Let the circumcircle of
be
. Then, there is a point
and a point
on circle
such that
and
trisects
and
, and point
lies on minor arc
. Point
is chosen on segment
such that
is one of the altitudes of
. Ray
intersects
at point
(not
) and is extended past
to point
, and
. Point
is also on
and
. Let the perpendicular bisector of
and
intersect at
. Let
be a point such that
is both equal to
(in length) and is perpendicular to
and
is on the same side of
as
. Let
be the reflection of point
over line
. There exist a circle
centered at
and tangent to
at point
.
intersect
at
. Now suppose
intersects
at one distinct point, and
, and
are collinear. If
, then
can be expressed in the form
, where
and
are not divisible by the squares of any prime. Find
.
Someone mind making a diagram for this?
User Counts
If this is you first time visiting this page, change the number below by one. (Add 1, do NOT subtract 1)
Doesn't that look like a number on a pyramid?
Answer key & solution to the problems
I will leave a big gap below this sentence so you won't see the answers accidentally.
dsf
fsd
Here:
1. 049
2. 170
3. 736
4. 011
5. 054
Solutions:
Problem 1
There is one and only one perfect square in the form
Find that perfect square.
Solutions
.
Suppose
.
Then,
, so since
,
so
is less than both
and
and thus we have
and
. Adding them gives
in some order. Hence,
.