problem solving

[stephanson12]

We call a positive integer n peculiar if, for any positive divisor d of n, the integerd(d + 1) divides n(n + 1). Prove that for any four different peculiar positive integers A, B, C and D,the following holds:
gcd(A, B, C, D) = 1.
Here gcd(A, B, C, D) is the largest positive integer that divides all of A, B, C and D.

hello, can someone help me please

 

[The Highlander]

We call a positive integer n peculiar if, for any positive divisor d of n, the integerd(d + 1) divides n(n + 1). Prove that for any four different peculiar positive integers A, B, C and D,the following holds:
gcd(A, B, C, D) = 1.
Here gcd(A, B, C, D) is the largest positive integer that divides all of A, B, C and D.

hello, can someone help me please

I presume you meant to write:-

We call a positive integer n peculiar if, for any positive integer divisor of n, d, the integer d(d + 1) divides n(n + 1) evenly.
Prove that for any four different peculiar positive integers A, B, C and D, the following holds:-

gcd(A, B, C, D) = 1.​

Here gcd(A, B, C, D) is the largest positive integer that divides all of A, B, C and D.

Please confirm that is the case or correct it and then show us what you have tried so far and tell us exactly where you are stuck.

Please follow the rules of posting in this forum, as provided at:-

READ BEFORE POSTING​

 

[stephanson12]

I don't know how to use the following

We call a positive integer n peculiar if, for any positive integer divisor[of n, d, the integer d(d + 1) divides n(n + 1) evenly.
Prove that for any four different peculiar positive integers A, B, C and D, the following holds:-

gcd(A, B, C, D) = 1.​

Here gcd(A, B, C, D) is the largest positive integer that divides all of A, B, C and D.

 

[Dr.Peterson]

We call a positive integer n peculiar if, for any positive divisor d of n, the integerd(d + 1) divides n(n + 1). Prove that for any four different peculiar positive integers A, B, C and D,the following holds:
gcd(A, B, C, D) = 1.
Here gcd(A, B, C, D) is the largest positive integer that divides all of A, B, C and D.

hello, can someone help me please

I would start by experimenting, to get a feel for what it means. Can you find a "peculiar number"? Can you find four of them? (It's conceivable that they are very rare and hard to find, so you might not be able to do this; but just trying should help you think about the problem better.)

Let's try an example. Is n=6 a peculiar number? It has divisors d=1, 2, 3, and 6. Here n(n+1) = 6*7 = 42; and we want that to be divisible by 1(1+1) = 2, 2(2+1) = 6, 3(3+1) = 12 , and 6(6+1) = 42. We find that 2, 6, and 42 all divide 42, but 12 does not, so 6 is not peculiar.

Does this at least help in understanding what you are being asked? I haven't thought any further than this, so I don't know yet whether peculiar numbers even exist. But I can see how it might work if you pick a less (or more) special number than 6 ...

I have no idea what to suggest for a proof, unless you find that you can describe all peculiar numbers more simply. If you show us some thoughts of your own, we will probably have some more ideas for you.

 

[The Highlander]

I don't know how to use the following

All Prime numbers are peculiar. Can you see why?

But not all peculiar numbers are prime numbers.

For example: 15 & 95 are peculiar (but not prime; why?).

Does that help you get started?

(Hint: Examine the factors of the non-prime numbers I gave you.)

 

[stephanson12]

All Prime numbers are peculiar. Can you see why?

But not all peculiar numbers are prime numbers.

For example: 15 & 95 are peculiar (but not prime; why?).

Does that help you get started?

(Hint: Examine the factors of the non-prime numbers I gave you.)

ah thank you, I understand that all prime numbers are peculair. But I don't understand your example with 15 (=5*3) and 95 (=5*19). I know that these numbers are peculair, but I see no connection between the two numbers

 

[The Highlander]

ah thank you, I understand that all prime numbers are peculair. But I don't understand your example with 15 (=5*3) and 95 (=5*19). I know that these numbers are peculair, but I see no connection between the two numbers

What is it you are being asked to show?

And what, if anything, does it suggest to you?

I gave you two non-prime peculiar numbers: 15 & 95 (287 is a third) and suggested you investigate them further (see Hint).

You really need to show us some work that you have done before more help is likely to be forthcoming.

I also asked if you understood why all primes are peculiar. You could start by thinking about that and offering an explanation of it.

 

[BigBeachBanana]

The first 5 non-prime peculiars are 15, 95, 287, 1199, and 4607.
As an interest in this problem, here's an Excel formula that checks if a number is peculiar. Note this only works for version 365.

=LET(
s,A2/SEQUENCE(A2),
d,FILTER(s,MOD(s,1)=0),
dd,d*(d+1),n,A2*(A2+1),
m,MOD(n/dd,1)=0,SUM(--m)=ROWS(m))

This formula checks if the number is prime.

=IF(A2=2,TRUE,IF(AND(MOD(A2,ROW(INDIRECT("2:"&ROUNDUP(SQRT(A2),0)))) <>0),TRUE,FALSE))

 

[stephanson12]

All primes are peculiar.
If n=prime, d can be n or 1.
-if d=1: d(d+1)=2 and this always is a divisor of n(n+1)
-if d=n= d(d+1)=n(n+1) and this always is a divisor of n(n+1)

The first few peculiar non-primes are
15 = 3 x 5
95 = 5 x 19
287 = 7 x 41
1199 = 11 x 109
4607 = 17 x 271
....

 

[blamocur]

All primes are peculiar.
If n=prime, d can be n or 1.
-if d=1: d(d+1)=2 and this always is a divisor of n(n+1)
-if d=n= d(d+1)=n(n+1) and this always is a divisor of n(n+1)

The first few peculiar non-primes are
15 = 3 x 5
95 = 5 x 19
287 = 7 x 41
1199 = 11 x 109
4607 = 17 x 271
....

This is what I get too. Also, it looks like all non-prime peculiars can be represented as $n=p(p^2-p-1)$ where $p$ is prime. At least all the ones under 99,999 look this way (but I have no clue how to prove this) :

$$\begin{array}{rcrcr} 15 &=& 3 &\times& 5 \\ 95 &=& 5 &\times& 19\\ 287 &=& 7 &\times& 41\\ 1199 &=&11 &\times& 109\\ 4607 &=& 17 &\times& 271\\ 23519 &=& 29 &\times& 811\\ 28799 &=& 31 &\times& 929\\ \end{array}$$

 

[The Highlander]

If n=prime, d can be n or 1.
-if d=1: d(d+1)=2 and this always is a divisor of n(n+1) Why?
-if d=n= d(d+1)=n(n+1) and this always is a divisor of n(n+1)
The first few peculiar non-primes are
15 = 3 x 5
95 = 5 x 19
287 = 7 x 41
1199 = 11 x 109
4607 = 17 x 271
....

The blue part (above) is just a direct copy of Massist's post in Math Forums!

You appear to be just sitting back and waiting for others to post the solution (or the steps towards it) without making very much effort yourself!

You need to 'tidy up' the green part so that it proves that all primes are peculiar.

There's almost certainly enough information been provided (by others) to enable the original question to be answered.

How about you take what's been given and show us your attempt to manipulate that (and the little bit of work that you appear to have done) towards a solution?

 

[Dr.Peterson]

I'm often curious where an interesting problem comes from; the only other place I find this one (besides Math Forums) is here:

https://www.egmo.org/egmos/egmo13/paper-day1-English.pdf

Interesting that it's dated yesterday. I don't know if that adds anything to our knowledge, except that it requires only high school level knowledge.

I myself would have waited to say anything concrete until we saw even a little bit of work. I just hinted "... unless you find that you can describe all peculiar numbers more simply", having seen the fact about primes (but not whatever comes next).

 

[The Highlander]

I'm often curious where an interesting problem comes from; the only other place I find this one (besides Math Forums) is here:

https://www.egmo.org/egmos/egmo13/paper-day1-English.pdf

Interesting that it's dated yesterday. I don't know if that adds anything to our knowledge, except that it requires only high school level knowledge.

I myself would have waited to say anything concrete until we saw even a little bit of work. I just hinted "... unless you find that you can describe all peculiar numbers more simply", having seen the fact about primes (but not whatever comes next).

As far as I can see, s/he's been given everything s/he needs to answer the question asked but appears to be reluctant to submit any fresh ideas or thoughts (maybe s/he just doesn't have any or any real enthusiasm to put the effort in? ).

A meagre couple of lines that s/he was led to by the nose (but still haven't been properly framed) hardly demonstrate good use of all the contributions made so far.

 

[blamocur]

I'm often curious where an interesting problem comes from; the only other place I find this one (besides Math Forums) is here:

https://www.egmo.org/egmos/egmo13/paper-day1-English.pdf

This probably merits a separate thread in a different section, but: how much time does it make sense to spend helping with olympiad-grade problems? They, of course, can be very interesting, but most of them don't do much in terms of better understanding of math.

 

[stephanson12]

This is what I get too. Also, it looks like all non-prime peculiars can be represented as $[COLOR=rgb(184, 49, 47)][B]n=p(p^2-p-1)[/B][/COLOR]$ where $p$ is prime. At least all the ones under 99,999 look this way (but I have no clue how to prove this) :

this means that n=p^3-p^2-p=p(p+1)(p-1)-p^2 = n
but we need to prove this, but i don't knew how

this looks a little bit like it;
p|n and p(p+1)|n(n+1) is given
so p(p+1)*Q = n^2+n
p(p+1)*Q - n^2 = n

 

[BigBeachBanana]

My thought is to use proof by contradiction assuming $\gcd(A, B ,C ,D) = k >1$.

 

[fresh_42]

This is what I get too. Also, it looks like all non-prime peculiars can be represented as $n=p(p^2-p-1)$ where $p$ is prime. At least all the ones under 99,999 look this way (but I have no clue how to prove this) :

$$\begin{array}{rcrcr} 15 &=& 3 &\times& 5 \\ 95 &=& 5 &\times& 19\\ 287 &=& 7 &\times& 41\\ 1199 &=&11 &\times& 109\\ 4607 &=& 17 &\times& 271\\ 23519 &=& 29 &\times& 811\\ 28799 &=& 31 &\times& 929\\ \end{array}$$

The roots of $p^2-p-1$ are the golden ratio and its negative solution. This brings the Fibonacci numbers into the game.
And $n(n+1)=(p^2 - p - 1) (p - 1)^2 p(p + 1)$ in case $n=p(p^2-p-1),$ i.e. $n+1=(p-1)^2(p+1).$

 

[fresh_42]

This is what I get too. Also, it looks like all non-prime peculiars can be represented as $n=p(p^2-p-1)$ where $p$ is prime. At least all the ones under 99,999 look this way (but I have no clue how to prove this) :

$$\begin{array}{rcrcr} 15 &=& 3 &\times& 5 \\ 95 &=& 5 &\times& 19\\ 287 &=& 7 &\times& 41\\ 1199 &=&11 &\times& 109\\ 4607 &=& 17 &\times& 271\\ 23519 &=& 29 &\times& 811\\ 28799 &=& 31 &\times& 929\\ \end{array}$$

We call a positive integer $n$ peculiar if, for any positive divisor $d$ of $n$ , the integer $d(d + 1)$ divides $n(n + 1).$

Lemma: If $n$ is peculiar and has at most two prime divisors then $n\in \{p\, , \,p(p^2-p-1)\}$ and $p^2-p-1$ is prime.

Proof: If $p\,|\,n$ then $n=p\cdot m.$ In case $m=1$ we have $p(p+1)=n(n+1)\,|\,n(n+1)$ and $1\cdot (1+1)= 2\,|\,n(n+1)$ so primes are peculiar. Let us therefore assume that $n$ is not prime and $m>1.$

Let $m=q$ be prime and $n=p\cdot q$ a product of two primes. In case $p=q$ we get
$$\begin{array}{lll} p(p+1)\,|\,n(n+1)&=p^2(p^2+1)=p^4+p^2=p^3(p+1)-p^3+p^2\\ &=p^3(p+1)-p^2(p+1)+2p^2\\ &=p^3(p+1)-p^2(p+1)+2p(p+1)-2p\\ &=p^3(p+1)-p^2(p+1)+2p(p+1)-2(p+1)+2\\ &\Rightarrow 0\equiv 2\bmod{p+1} \end{array}$$which is impossible since $p>1.$ Therefore $p\neq q$ and
$$\begin{array}{lll} q(q+1)\,|\,n(n+1)=pq(pq+1)&\Longrightarrow (q+1)\,|\,p(pq+1)=p^2(q+1)-p^2+p\\ &\Longrightarrow (q+1)\,|\,p(p-1) \end{array}$$and for symmetry reasons also $(p+1)\,|\,q(q-1).$ Thus $r\cdot(q+1)=p(p-1)$ and $s\cdot(p+1)=q(q-1)$ for some positive integers $r,s\in \mathbb{Z}.$ If $p=2$ then $(q+1)\,|\,2\cdot(2-1)=2$ which is impossible for $q>1.$ We may thus assume that we have odd primes $3\leq p <q$ and $q\,\nmid\,(p+1).$ Hence $q\,|\,s$ or $q\cdot t=s$ and $t\cdot(p+1)=q-1$ for some positive integer $t\in \mathbb{Z}.$

If $p\,\nmid\,(q+1)$ then $p\,|\,r,$ say $p \cdot u=r.$ Thus
$$0<u=\dfrac{p-1}{q+1}=\dfrac{p-1}{tp+t+2}\in \mathbb{Z}$$which implies first $t=0$ and then $s=0$ which is a contradiction. We therefore have $p\,|\,(q+1),$ say $p\cdot k=q+1,$ i.e. $r(q+1)=rpk=p(p-1)$ for some positive integer $k\in \mathbb{Z}$ and
$$p=rk+1 \text{ and }q=t(rk+2)+1=rk^2+k-1 .$$If $t\geq k$ then $rk^2+k-1=t(rk+2)+1\geq rk^2+ 2k+1$ which isn't possible for positive $k.$ Hence we may assume $t<k$ or $-k<-t.$ Now
$$\begin{array}{lll} 0&=rkt+2t-rk^2-k+2=r(kt-k^2)-(k-2t-2)\\ r&=\dfrac{k-2t-2}{k(t-k)}=\dfrac{2t+2-k}{k(k-t)}>0\\ 0&< 2t+2-k\text{ and thus }t<k<2(t+1)\\ k&\in \{t+1,\ldots,2t+1\}\, , \,k-t\in \{1,\ldots,t+1\}\\ r&=\dfrac{2t+2-k}{k(k-t)}\geq 1\\ 0&<k(k-t)\leq 2t+2-k < t+2< k+2\\ 0&<k-t< 1+\dfrac{2}{k} \end{array}$$Since both, $t$ and $k$ are positive, and $0<t<k,$ the smallest possible number is $k=2.$ On the other hand, if $k\geq 2,$ then $k-t \in ]0,2[.$ and so $k=t+1.$ We have then in all cases
$$r=\dfrac{2t+2-k}{k(k-t)}=\dfrac{2t+2-t-1}{t+1}=1$$and thus
$$q=p(p-1)-1=p^2-p-1 \text{ prime. }$$as demanded.

So if I didn't make any mistakes, this reduced the problem to ...

If $n$ is peculiar then it has at most $2$ prime factors.​

The original problem should be easier once we know what peculiar numbers look like.

 

[fresh_42]

Note: A number $n$ is peculiar if $d\,|\,n\Longrightarrow d(d+1)\,|\,n(n+1).$

I can prove the original statement in post #1 (four different peculiar numbers are coprime) and the general conjecture from post #10 (a peculiar number looks like a product of two primes $n=p\cdot q$ where $q=p^2-p-1$
in case $n$ has at most two prime factors, but I have no idea how to prove that.

Does anybody else have an idea how to prove that peculiar numbers cannot have 3 different divisors?

 

[BigBeachBanana]

Solution posted. See page 10.