Alternative (1) Solution 526:
We know that and we know that is irrational. This means taht we need only prove that is irrational.
We can do so easily, since . Now assume that this equals where then we have so that , contradiction so is irrational and we are done.

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A little basic number theory pilfered from Putnam.
Problem 537: **
Show that for each , where is prime.
.
Problem 538: **
Show that there are no integer solutions to:
where .Last edited by joostan; 31102015 at 00:44. 
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 31102015 00:42
(Original post by joostan)
A little basic number theory pilfered from Putnam.
Problem 537: **
Show that for each :
.
Problem 538: **
Show that there are no integer solutions to:
where . 
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 31102015 00:43
(Original post by Renzhi10122)
What's p? A prime?
Probably shoulda specified, but I'd say it's reasonably clear from context. 
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 31102015 00:57
(Original post by joostan)
Yarp, p's a prime.
Probably shoulda specified, but I'd say it's reasonably clear from context.
Solution 537
Firstly note that . By the LTE lemma, we have
, where is the greatest power of 2 that divides . Therefore, is divisible by , a stronger result. Now note that is not divisible by 3,5, or 17, since it is prime. By Fermat's Little Theorem, . Noting that each of 3, 5 and 17 can be written in the form , in particular, for , setting as each of the three primes, we see that 32 is a multiple of , and so is divisible by 3, 5 and 17.
Thus, and indeed for all primes .
EDIT: We could've used difference of two squares for the first bit, but only thought of that afterwardsLast edited by Renzhi10122; 31102015 at 01:01. 
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Solution 537 (2)
Damn I really was a bit slow :)Spoiler:Show
First note that
If p is a prime greater than 17 then p is odd and not divisible by 3, hence therefore and hence , or , in which case since and hence So .
Note that we can rewrite the expression as . Each factor is even since p is odd and hence any power of p is odd. Hence
Finally let us consider the divisor 17. If p is a prime greater than 17 then p and 17 are coprime. Hence by Fermat's Little Theorem
and hence so
So 3, 5, 64 and 17 divide . Hence 16320 divides .
Last edited by 16Characters....; 31102015 at 01:33. 
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 31102015 10:55
(Original post by Renzhi10122)
x(Original post by 16Characters....)
x
Protip: To typeset use a \equiv b \pmod c instead of a \equiv b (mod c) since that just squishes the variables together.
Edit: It evens saves you the hassle of putting the brackets around the mod and it just requires a slash and a p before the mod.Last edited by Zacken; 31102015 at 10:56. 
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Some more analysis
Problem 539
Let be any function that is continuous on
Prove that there exists a constant such that
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(Original post by Indeterminate)
Some more analysis
Problem 539
Let be any function that is continuous on
Prove that there exists a constant such that
Posted from TSR Mobile
Suppose that for all . Then we must have , which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.
Thus there must exist for which , and such that . Suppose without loss of generality we have . Let . Then at a, and at b . Thus by the intermediate value theorem, there exists a such that for which , which implies by construction of g
Edit: A nice intuitive picture for how this theorem works: Draw a square, and from the bottom left corner to the top right corner, draw a curve resembling the graph of sin from 0 to pi/2. Now try and join the left side to the right side with a continuous curve that doesn't intersect the sine curve .Last edited by FireGarden; 02112015 at 02:00. 
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(Original post by FireGarden)
Solution 539.
Suppose that for all . Then we must have , which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.
Thus there must exist for which , and such that . Suppose without loss of generality we have . Let . Then at a, and at b . Thus by the intermediate value theorem, there exists a such that for which , which implies by construction of g
Define g(x) = sin(pi x / 2)  f(x). Then g(0) = f(0), g(1) = 1 f(1) and so since we know 0<=f(x)<=1, g(0)<=0, g(1)>=0. If we have equality in either case we're done, otherwise by IVT we can find k in (0,1) with g(k) = 0 and we're done.
[Questions like this come up often enough that you should make sure you know how to answer them concisely. The defn of IVT I use requires f(a) < 0, f(b) > 0 (i.e. it doesn't allow equality) or it could go a little shorter still]. 
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Problem 540 (**/***)
a) Show that:
stating any conditions that must hold for this to be valid.
b) Find the value of
(Hints available if necessary) 
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(Original post by atsruser)
x
I can't quite wrap my head around the whole thing. 
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 05112015 15:50
Can't believe I've never used this thread before.
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UProblem 541
I am on a IPad so not really sure about this bold stuff etc.
Prove the for natural n the fraction (21n+4)/(14n+3) is irreducible.
I will post a solution later on today.
This is a problem from an old IMO problem!
Posted from TSR MobileLast edited by physicsmaths; 05112015 at 18:07. 
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Solution 541:
Observe that .
Thus if and for then so .
Hence we have and coprime, so:
is irreducible. 
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(Original post by joostan)
Solution 541:
Observe that .
Thus if and for then so .
Hence we have and coprime, so:
is irreducible.
Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
=gcd(7n+1,14n+32(7n+1))
=gcd(7n+1,1)=1 Q.E.D
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(Original post by physicsmaths)
Mine essentially the same but slight different at the start I think.
Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
=gcd(7n+1,14n+32(7n+1))
=gcd(7n+1,1)=1 Q.E.D
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Also  it's fine to post a solution to your own problem once somebody else has already answered it, but not before.Last edited by Zacken; 05112015 at 17:55. 
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(Original post by 16Characters....)
Problem 508
is an N x N matrix with N distinct eigenvalues all in the range . Prove that:
Converges. {NB: I denotes the N x N identity matrix}
Spoiler:Show
A matrix with all of its eigenvalues distinct is diagonalisable.

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(Original post by 16Characters....)
Problem 508
is an N x N matrix with N distinct eigenvalues all in the range . Prove that:
Converges. {NB: I denotes the N x N identity matrix}
Since has distinct eigenvalues, we can write for some fixed, invertible matrix , and where .
Notice that:
.
But .
(Assuming , if so then so still converges.)
Hence we have that:
.
Thus the sum converges.Last edited by joostan; 05112015 at 21:21.
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