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# The Proof is Trivial! Watch

1. (Original post by bananarama2)
LOTF Solution 336

By maths we have the inequality.
I'm still not sure if he actually found a solution, or just put up some random statements int eh hope that nobody would check
2. (Original post by james22)
I'm still not sure if he actually found a solution, or just put up some random statements int eh hope that nobody would check
Just notice that from AM-GM, .
The second inequality is just an application of .

Problem 337**

Let and be externally tangent circles at and the radius of is greater than the radius of . Let be a point on which does not lie on the line joining the centres of and . Let and be points on such that and are tangent to . The lines and intersect again at and , respectively. Let be the point of intersection of the tangent to at and the line . Prove that as varies, the locus of is a line.
3. Problem 338*

Consider the sequence for , what is the maximum value of ?
4. Solution 338

Note . Now suppose for there is s.t. and then . Since we have
5. (Original post by bananarama2)
LOTF Solution 336

By maths we have the inequality.
HA! How unfair, I even said AM-GM and everything.
6. Two warm-up problems.

Problem 339**

Let be a sequence of non-negative integers. Let where . Prove that for any the number of integers such that is .

Problem 340**

Find all maps such that for all the relation holds. Now let be an arbitrary positive integer. Find all maps such that for all choices of positive integers .
Two warm-up problems.

Problem 339**

Let be a sequence of non-negative integers. Let where . Prove that for any the number of integers such that is .

Problem 340**

Find all maps such that for all the relation holds. Now let be an arbitrary positive integer. Find all maps such that for all choices of positive integers .
Warm up? Lol

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8. Solution 340** (partial)

Attention everyone! I have made unprecedented progress with this fiendish problem. I have found one of the three possible maps - now I am just in a quandary as to what the other two could possibly be?

UPDATE!
Spoiler:
Show
Eureka, I HAVE FOUND THE SECOND MAP! It is none other than the sentient map from Dora the Explorer! Now for the final map... which continues to antagonize and elude me. I shall update this post when I find that tertiary map and bring it to justice in the form of a fully worked solution to the problem stated.

Problem 340**

Find all maps such that for all the relation holds. Now let be an arbitrary positive integer. Find all maps such that for all choices of positive integers .
Ooh, I like this problem, although my solution is a bit plodding and slow. So plodding and slow, in fact, that I have only done the first part. The induction I used is very messy - I'd appreciate it if someone could check it.
Spoiler:
Show

Setting m=n=1 we have 2 f(1) | 2, and hence f(1) = 1.
Then m=1 gives 1+f(n) | n+1 for all n.
Also setting m=n we have 2 f(m) | 2m, and so f(m) | m, for all m.

f(2)+f(n) | 2+n
Let n be prime > 2. Suppose f(n) = n. Then f(2)+f(n) = f(2)+n | 2+n, but f(2) is 1 or 2, so we must have f(2) = 2. Suppose f(n) = 1. Then f(2)+1 | 2+n, and hence we can't have f(2) = 1 (because then even divides odd). So f(2) = 2.

Now let m,n be odd prime, with m<n. Then f(m)+f(n) | m+n. Suppose f(n) = n. Then f(n) is greater than half m+n, and hence f(m)+f(n) = m+n, and so f(m) = m. So if we have one odd prime for which f(p) = p, then all lower odd primes have f(q) = q. In other words, considering the odd primes, we must have f(p_i) = p_i for i=1 to k, and then f(p_i) = 1 for i>k (where k might be infinity).

Now consider p prime, such that f(p) = 1. Then f(m)+1 | m+p, for all m. Letting m=2, we have that 3|p+2. So p must be 1 mod 3 in order for f(p) to be 1. But because we know that there is an infinite number of primes 1 mod 3 and 2 mod 3, and using the previous paragraph, we then see that f(p) = p for all primes p.

To summarise, so far we have that f(n) = n for n prime and for n=1.

Unnecessary further result:
Spoiler:
Show

Now, f(p - n)+f(n) | p, for all n<p and for all prime p. So f(p-n)+f(n) = 1 (contradiction because the LHS is greater than or equal to 2), or f(p-q)+f(q) = p, for all q<p and prime p. Hence f(p-q) = p-q for any primes q < p. This is a very powerful statement - any number that is prime or is a difference of two primes, has f(n) = n. This reasoning works not just for prime q, but for any q where f(q) = q. So, in particular, f(p-1) = p-1.

Then, f(qp-q)+f(q) | qp, so f(qp-q)+q | qp, and dividing by q yields f(q(p-1)) = q(p-1), for prime p,q.

Now we proceed inductively. Base case: p=2, where all f(2p-n) have f(2p-n) = 2p-n.
Then assume that for p=p, we have that all f(2p-n) = 2p-n. Then considering q the next prime up from p, we have that f(2q-n)+f(n) | 2q. We only need to check the n that will give us 2p <= 2q-n < 2q, because the n that will give 2q-n < 2p have already been checked under the inductive hypothesis. Hence n <= 2(q-p). But Bertrand's Postulate states that q<2p, so n < 2(2p-p) = 2p. And we have [inductive hypothesis] that all such n have f(n) = n, so f(2q-n) + n | 2q. Hence f(2q-n)+n = 2, q or 2q.
If it's 2q, we're done straight away.
It can't be 2, because that requires n=1, f(2q-1) = 1, but then f(2q-1)+f(q+1) | 3q and 1+f(q+1) | 3q - but we know that since q+1<2p [strong form of Bertrand], we have f(q+1) = q+1, and q+2 | 3q which is rubbish.
Suppose that f(2q-n)+n = q. Then f(2q-n) = q-n; then q-n | 2q-n. But then there is k such that k(q-n) = 2q-n, from which (k-2)q = (k-1)n. If k is odd, then the LHS is odd and the RHS even; if k is even, then the LHS is even so n is even. Hence q-n | 2(q-n/2), but q-n is odd and so q-n | q-n/2, which is a contradiction.
Therefore, f(2q-n) = 2q-n for all 1 <= n < 2q.

So by extremely messy induction, we're done: the only valid map is f(n) = n for all n.
10. (Original post by Smaug123)
Spoiler:
Show
Ooh, I like this problem, although my solution is a bit plodding and slow. So plodding and slow, in fact, that I have only done the first part. The induction I used is very messy - I'd appreciate it if someone could check it.
Spoiler:
Show

Setting m=n=1 we have 2 f(1) | 2, and hence f(1) = 1.
Then m=1 gives 1+f(n) | n+1 for all n.
Also setting m=n we have 2 f(m) | 2m, and so f(m) | m, for all m.

f(2)+f(n) | 2+n
Let n be prime > 2. Suppose f(n) = n. Then f(2)+f(n) = f(2)+n | 2+n, but f(2) is 1 or 2, so we must have f(2) = 2. Suppose f(n) = 1. Then f(2)+1 | 2+n, and hence we can't have f(2) = 1 (because then even divides odd). So f(2) = 2.

Now let m,n be odd prime, with m<n. Then f(m)+f(n) | m+n. Suppose f(n) = n. Then f(n) is greater than half m+n, and hence f(m)+f(n) = m+n, and so f(m) = m. So if we have one odd prime for which f(p) = p, then all lower odd primes have f(q) = q. In other words, considering the odd primes, we must have f(p_i) = p_i for i=1 to k, and then f(p_i) = 1 for i>k (where k might be infinity).

Now consider p prime, such that f(p) = 1. Then f(m)+1 | m+p, for all m. Letting m=2, we have that 3|p+2. So p must be 1 mod 3 in order for f(p) to be 1. But because we know that there is an infinite number of primes 1 mod 3 and 2 mod 3, and using the previous paragraph, we then see that f(p) = p for all primes p.

To summarise, so far we have that f(n) = n for n prime and for n=1.

Unnecessary further result:
Spoiler:
Show

Now, f(p - n)+f(n) | p, for all n<p and for all prime p. So f(p-n)+f(n) = 1 (contradiction because the LHS is greater than or equal to 2), or f(p-q)+f(q) = p, for all q<p and prime p. Hence f(p-q) = p-q for any primes q < p. This is a very powerful statement - any number that is prime or is a difference of two primes, has f(n) = n. This reasoning works not just for prime q, but for any q where f(q) = q. So, in particular, f(p-1) = p-1.

Then, f(qp-q)+f(q) | qp, so f(qp-q)+q | qp, and dividing by q yields f(q(p-1)) = q(p-1), for prime p,q.

Now we proceed inductively. Base case: p=2, where all f(2p-n) have f(2p-n) = 2p-n.
Then assume that for p=p, we have that all f(2p-n) = 2p-n. Then considering q the next prime up from p, we have that f(2q-n)+f(n) | 2q. We only need to check the n that will give us 2p <= 2q-n < 2q, because the n that will give 2q-n < 2p have already been checked under the inductive hypothesis. Hence n <= 2(q-p). But Bertrand's Postulate states that q<2p, so n < 2(2p-p) = 2p. And we have [inductive hypothesis] that all such n have f(n) = n, so f(2q-n) + n | 2q. Hence f(2q-n)+n = 2, q or 2q.
If it's 2q, we're done straight away.
It can't be 2, because that requires n=1, f(2q-1) = 1, but then f(2q-1)+f(q+1) | 3q and 1+f(q+1) | 3q - but we know that since q+1<2p [strong form of Bertrand], we have f(q+1) = q+1, and q+2 | 3q which is rubbish.
Suppose that f(2q-n)+n = q. Then f(2q-n) = q-n; then q-n | 2q-n. But then there is k such that k(q-n) = 2q-n, from which (k-2)q = (k-1)n. If k is odd, then the LHS is odd and the RHS even; if k is even, then the LHS is even so n is even. Hence q-n | 2(q-n/2), but q-n is odd and so q-n | q-n/2, which is a contradiction.
Therefore, f(2q-n) = 2q-n for all 1 <= n < 2q.

So by extremely messy induction, we're done: the only valid map is f(n) = n for all n.
It is fine; however, you could have eluded this induction.
Spoiler:
Show
From for all primes , we get , which then gives . Now, choosing sufficiently large, we are done.
11. (Original post by Pterodactyl)
Spoiler:
Show
I think this is correct.
Solution 340
Part I
We have two cases to consider
If
for for any . When m=n this only occurs for k = 1, m. So f(x) = 1, f(x) = x

Now to check that these work for the other case
If is not always even so
If , which clearly divides so the only map is

Part II
Using the same cases again.
If
for
So where

Now to check if these work for the other case.

and
Clearly will not always be a multiple of some power of 2 so .
is the only mapping, again.

You actually have , unless you prove that is constant.
The second part is also incorrect. The same mistake plus not all divisors of can be written as .
You actually have , unless you prove that is constant.
The second part is also incorrect. The same mistake plus not all divisors of can be written as .
Ah, stupid of me! Deleted the post.
It is fine; however, you could have eluded this induction.
Spoiler:
Show
From for all primes , we get , which then gives . Now, choosing sufficiently large, we are done.
Aah, much better - thanks!
14. Problem 341**/***

Find

I don't actually know how to solve this one, but I do have the answer.
15. (Original post by james22)
Problem 341**/***

Find

I don't actually know how to solve this one, but I do have the answer.
Solution 158. Notice that the function is even.
Solution 158. Notice that the function is even.
Completely separate note to the maths; I love your really broad use of the English languages "Thence."
17. This is an easier one than the normal problems here, maybe for those who don't look at this thread very often.

Problem 342*

Evaluate
18. Problem 343 */**

Prove that:

19. Solution 343

Let denote the proposition that

Consider the case when .

Assume to be true, and consider the case for :

Hence is true if is assumed to be true which, along with , implies that is true by mathematical induction.
20. (Original post by Ateo)
This is an easier one than the normal problems here, maybe for those who don't look at this thread very often.

Problem 342*

Evaluate
Solution 342

Problem 344*/***

Evaluate .

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