09 Jun 09
i could be wrong though, and would LOVE to see a direct proof! 🙂Here goes!
If m=1 then m is a perfect square.
Assuming m is not 1 then there is a prime-factor P^(2a+1) dividing m.
Since we know that m^2 + n^2 + m = bmn (for some positive integer b), it is clear that m divides n^2. Since m divides n^2; and since P^(2a+1) divides m; it follows that P^(2a+1) divides n^2.
But, it further is the case that P^(2a+2) divides n^2 because n^2 is a perfect square. It follows that P^(a+1) divides n. Of course, P^(a+1) also divides m.
Since P^(a+1) divides both m and n, it follows that P^(2a+2) divides mn. Since bmn = m^2 + n^2 + m, it follows that P^(2a+2) divides m^2 + n^2 + m.
Since P^(2a+2) divides both m^2 and n^2, it follows that P^(2a+2) also divides m.
So if m is divisible by an odd power of some primenumber, it is also divisible by an even power of that prime number.
So m is a perfect square.
10 Jun 09
Sorry guys! I've been cavorting around in other forums lately. Excellent work by LemonJello and TheMaster37! Like I said, I don't have any answers to these questions, but the proofs presented are quite convincing.
Next problem!
Problem 5. There are several points on the plane. It is given that any four of them create a convex quadrilateral. Prove that all of them create a convex polygon.
22 Jun 09
Originally posted by PBE6Trying to find a direct proof again, a proof with contradiction isn't that hard 🙂
Sorry guys! I've been cavorting around in other forums lately. Excellent work by LemonJello and TheMaster37! Like I said, I don't have any answers to these questions, but the proofs presented are quite convincing.
Next problem!
Problem 5. There are several points on the plane. It is given that any four of them create a convex quadrilateral. Prove that all of them create a convex polygon.
02 Jul 09
1 edit
Originally posted by blacknight1985the Pooh & Piglet question as posed could have been read either way.
pls take a look at my first post..u hav misread the question..😀
actually, I like the first reading better as it makes it into a "trick question" where lots of people would waste a lot of time making calculations to find out low long it took Pooh to get to Piglet's house. The trick is that you need only to establish that Pooh is traveling faster than Piglet.
If Pooh and Piglet were traveling at the same rate, after meeting, each would have had 2 minutes to go before reaching the other's place. But Pooh only has 1 minute left to go when they meet - so Pooh must be past the halfway point. So Pooh was traveling faster - as a result, there's no need to calculate how long it took Pooh to get there - it's clearly less than the 4 minutes it took Piglet to travel -- so the answer would be 4 minutes.
05 Jul 09
Originally posted by PBE6(0,0)
Sorry guys! I've been cavorting around in other forums lately. Excellent work by LemonJello and TheMaster37! Like I said, I don't have any answers to these questions, but the proofs presented are quite convincing.
Next problem!
Problem 5. There are several points on the plane. It is given that any four of them create a convex quadrilateral. Prove that all of them create a convex polygon.
(0,2)
(2,0)
(2,2)
Coplanar, form a convex quadrilateral (square). If the fifth point is (1,1), you do not get a convex polygon. It cannot be drawn with those five vertices. Thus, this has been disproved by counterexample.
05 Jul 09
Originally posted by AThousandYoungSeems to me that (0,0)-(0,2)-(1,1)-(2,0) is a trilateral, not a quadrilateral.
(0,0)
(0,2)
(2,0)
(2,2)
Coplanar, form a convex quadrilateral (square). If the fifth point is (1,1), you do not get a convex polygon. It cannot be drawn with those five vertices. Thus, this has been disproved by counterexample.
08 Jul 09
No luck without contradictions. Here is A proof.
The given state that any four points form a convex quadrilateral, from which we know that no three points are on one straight line.
Let P be the convex hul of all points.
Suppose there is a point X inside P (not on the edge).
Then there are three points (say A, B and C) so that X is inside ABC because no three points are on one straight line.
But then the quadrilateral ABCX is not convex!
Contradiction.
So all points lie on the edge of the convex hull P, making P a convex polyhedron.
15 Aug 09
1 edit
Originally posted by steven9It should be equal to 25*26/2 = 325. In general, if you add up the numbers from 1 to m the result should be m*(m+1)/2.
Hi all add up all the numbers from 1 to 25. What is the total.
Here's a quick sketch of one way to prove it: Let S(m) denote the sum we are looking for.
Then [S(m)] + [S(m)] = [1 + 2 + 3 + ... + (m-1) + m] + [m + (m-1) + (m-2) + ... + 2 + 1].
So in the first [S(m)] I have written the sum from 1 to m, whereas in the second [S(m)] I have written it from m to 1. This is just to see that you can then pair up corresponding entries from the first [S(m)] and from the second [S(m)] to get:
S(m) + S(m) = (m+1) + (m+1) + (m+1) + ... + (m+1) + (m+1)
=> 2*S(m) = m*(m+1)
So, S(m) = m*(m+1)/2.
25 Aug 09
Originally posted by LemonJelloIt certainly is a problem...
Here's another math problem.
Find all prime numbers, p, such that (p-1)! + 1 is a power of p.
For example, (2-1)! + 1 = 2^1. So 2 is one such prime. Find all the others.
First we can state the problem as:
(p-1)!+1 = p^k for some k.
assume k=1.
(p-1)!+1=p
=> (p-1)!=p-1
which has only the trivial solution of p=2.
Now assume k=2
(p-1)!+1=p^2
Primes for which (p-1)!+1 = 0 mod p^2 are called Wilson Primes, only 3 are known:
5,13 and 563 with no others below 5*10^8
source: http://mathworld.wolfram.com/WilsonPrime.html
as 12!+1 > 13^2 and 562!+1>563^2
Since 4!+1 = 5^2, 5 is a solution.
now assume k>2
(p-1)!+1=p^k implies (p-1)!+1 = 0 mod p^2 so again, only 5,13,563 could possibly be solutions less than 5*10^8.
Now a quick computation shows:
12!+1 <> 0 mod 13^3
=> 12!+1 <> 0 mod 13^k for all k>=3
and
562!+1 <> 0 mod 563^3
=> 562!+1 <> 0 mod 563^k for all k>=3.
Hence the only known solutions are p=2 and p=5.
Certainly there are no others with p < 5*10^8.
It is conjectured however that there are an infinite number of Wilson primes