Why not 12? (about semisimple numbers)

[Yarem4uk]

This problem consists of three parts. I solved 1st and 2nd. But 3th...
1) Let x = 3p , x+1=2q, x+N= 3r , x+N+1=2s, where p , q , r , s are some simple numbers and N is some natural number. What the smallest value can get N?
2) Let y = 2p , y+1=3q, y+N= 2r , y+N+1=3s, where p , q , r , s are some simple numbers and N is some natural number. Prove that N>=12.
3) Is there such a number y, that N=12?

1) and 2) problems are not difficult. For them N>=12 and N=12k, k=1,2,3... I wrote a program to check the correctness of the decision, when N=12. For the first problem I got, for example, such results
q=5, s=11, p=3, r=7, x=9
q=11, s=17, p=7, r=11, x=21
q=101, s=107, p=67, r=71, x=201
...
But there are no results for the 2nd problem, when N=12. Why? I can not understand. So the answer to the third question is "no"?
I decided to take N=24 for the 2nd problem, and got such results
q=5, s=13, p=7, r=19, y=14
q=101, s=109, p=151, r=163, y=302
...
What's wrong with the number N=12? Please help me and sorry for my mistakes. Perhaps, this is a problem for the topic "Advanced Math".

 

[Dr.Peterson]

This problem consists of three parts. I solved 1st and 2nd. But 3th...
1) Let x = 3p , x+1=2q, x+N= 3r , x+N+1=2s, where p , q , r , s are some simple numbers and N is some natural number. What the smallest value can get N?
2) Let y = 2p , y+1=3q, y+N= 2r , y+N+1=3s, where p , q , r , s are some simple numbers and N is some natural number. Prove that N>=12.
3) Is there such a number y, that N=12?

1) and 2) problems are not difficult. For them N>=12 and N=12k, k=1,2,3... I wrote a program to check the correctness of the decision, when N=12. For the first problem I got, for example, such results
q=5, s=11, p=3, r=7, x=9
q=11, s=17, p=7, r=11, x=21
q=101, s=107, p=67, r=71, x=201
...
But there are no results for the 2nd problem, when N=12. Why? I can not understand. So the answer to the third question is "no"?
I decided to take N=24 for the 2nd problem, and got such results
q=5, s=13, p=7, r=19, y=14
q=101, s=109, p=151, r=163, y=302
...
What's wrong with the number N=12? Please help me and sorry for my mistakes. Perhaps, this is a problem for the topic "Advanced Math".

What is a simple number? Do you perhaps mean prime?

What have you been learning that this is intended to test? I would put it in the category of number theory. What theorems have you learned that are relevant? How did you solve the first two parts? It's very helpful when we can see what you know.

 

[Yarem4uk]

What is a simple number? Do you perhaps mean prime?

What have you been learning that this is intended to test? I would put it in the category of number theory. What theorems have you learned that are relevant? How did you solve the first two parts? It's very helpful when we can see what you know.

Thank you!


Yes, prime and in the category of number theory . My english is bad, sorry.

My solution for the first two parts
1)
x + N = 3r and x = 3p => 3p+N=3r => N : 3
x + N + 1 = 2s and x+1 = 2q => 2q+N=2s => N : 2
So, N=6k, k=1,2,3...
Let N=6. Than we have from x+N+1=2s
x+1+6=2s|:2
q+3=s but it is impossible when q and s are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12
2) similarly for the part two

y + N = 2r and y = 2p => 2p+N=2r => N : 2
y + N + 1 = 3s and y+1 = 3q => 3q+N=3s => N : 3
So, N=6k, k=1,2,3...
Let N=6. Than we have from y+N=2r
y+6=2r|:2
p+3=r but it is impossible when p and r are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12
But, i can not find for the N=12 such y...

 

[Dr.Peterson]

Thank you!
Yes, prime and in the category of number theory. My english is bad, sorry.

My solution for the first two parts
1)
x + N = 3r and x = 3p =>; 3p+N=3r => N : 3
x + N + 1 = 2s and x+1 = 2q => 2q+N=2s => N : 2
So, N=6k, k=1,2,3...
Let N=6. Than we have from x+N+1=2s
x+1+6=2s|:2
q+3=s but it is impossible when q and s are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12
2) similarly for the part two
y + N = 2r and y = 2p => 2p+N=2r => N : 2
y + N + 1 = 3s and y+1 = 3q => 3q+N=3s => N : 3
So, N=6k, k=1,2,3...
Let N=6. Than we have from y+N=2r
y+6=2r|:2
p+3=r but it is impossible when p and r are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12
But, I can not find for the N=12 such y...

I have not looked into this further yet, but for the sake of others reading this, I want to suggest that you are using ":" to mean "divided by" (which is standard in many countries), but here you are using N:2 to mean that N is divisible by 2, which in my experience is properly denoted by "2|N". Is that notation not used in your region?

But I don't know what you mean by "x+1+6=2s|:2".

 

[Yarem4uk]

I have not looked into this further yet, but for the sake of others reading this, I want to suggest that you are using ":" to mean "divided by" (which is standard in many countries), but here you are using N:2 to mean that N is divisible by 2, which in my experience is properly denoted by "2|N". Is that notation not used in your region?

But I don't know what you mean by "x+1+6=2s|:2".

We use three vertical dot to mean "divided by" and I don't know how to replace it. Perhaps, it's better to just write "divided by".
And "x+1+6=2s|:2" means that all equation need to be divided on 2.

 

[Dr.Peterson]

A couple more comments:

My solution for the first two parts
1)
x + N = 3r and x = 3p =>; 3p+N=3r => N : 3
x + N + 1 = 2s and x+1 = 2q => 2q+N=2s => N : 2
So, N=6k, k=1,2,3...
Let N=6. Than we have from x+N+1=2s
x+1+6=2s|:2

I think what you mean here is "divide the equation by 2" (after replacing x+1 with 2q).

q+3=s but it is impossible when q and s are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12

I think you overstated this. q+3=s is impossible for two odd primes; but it is possible if q=2 and s=5! And, in fact, this gives a solution. So in fact N can be 6. You will find that x=3, so that 3|x, 2|(x+1), 3|(x+6), and 2|(x+7). Of course, here p=1, which is not prime (by modern definitions --at one time it was called prime!), so this is not a solution (unless your term "simple" is not identical to "prime", and includes 1). But it is important not to overlook such details.

Also, note that if your work was correct, you showed only that N must be an even multiple of 6, not that every even multiple of 6 does actually yield a solution. To actually state that 12 is the smallest value of N, you have to show the solution for N=12. (Are you allowed to use a program to do this, or was that just intended as a check?) I presume this is no harder than my finding the solution for N=6, but it has to be stated explicitly.

2) similarly for the part two
y + N = 2r and y = 2p => 2p+N=2r => N : 2
y + N + 1 = 3s and y+1 = 3q => 3q+N=3s => N : 3
So, N=6k, k=1,2,3...
Let N=6. Than we have from y+N=2r
y+6=2r|:2
p+3=r but it is impossible when p and r are prime
It will be for all N=6k, when k is odd (1, 3, 5...)
So, N=12
But, I can not find for the N=12 such y...

Again, you have shown (if all your work is correct -- the same issue arises as above) only that N can't be less than 12, which is just what they asked you to do. There may well be other considerations that prevent it from being 12; I see nothing surprising in that, in itself.

The big question is, how to prove that it can't be 12, if that is true! What goes wrong when you try to find the solution? Do you see any possible reasons? (I haven't tried yet.)

 

[Yarem4uk]

Thank you very much for the comments!

I think what you mean here is "divide the equation by 2" (after replacing x+1 with 2q).

Yes, you're right

I think you overstated this. q+3=s is impossible for two odd primes; but it is possible if q=2 and s=5! And, in fact, this gives a solution. So in fact N can be 6. You will find that x=3, so that 3|x, 2|(x+1), 3|(x+6), and 2|(x+7). Of course, here p=1, which is not prime (by modern definitions --at one time it was called prime!), so this is not a solution (unless your term "simple" is not identical to "prime", and includes 1). But it is important not to overlook such details.

1 is not prime, yes

Also, note that if your work was correct, you showed only that N must be an even multiple of 6, not that every even multiple of 6 does actually yield a solution. To actually state that 12 is the smallest value of N, you have to show the solution for N=12. (Are you allowed to use a program to do this, or was that just intended as a check?) I presume this is no harder than my finding the solution for N=6, but it has to be stated explicitly.

For 12
1) x+12=3r|:3
p+4=r and It's possible
x+1+12=2s|:2
q+6=s and It's possible too
So, N>=12.
2) y+12=2r|:2
p+6=r and It's possible
y+1+12=3s|:3
q+4=s and It's possible too
So, N>=12 too.

Again, you have shown (if all your work is correct -- the same issue arises as above) only that N can't be less than 12, which is just what they asked you to do. There may well be other considerations that prevent it from being 12; I see nothing surprising in that, in itself.

The big question is, how to prove that it can't be 12, if that is true! What goes wrong when you try to find the solution? Do you see any possible reasons? (I haven't tried yet.)

So N>=12, and I found such x with a program help. For this I checked all combinations of prime numbers p, q, r, s if p, q, r, s < 1000000.
But for y (2nd task) the program did not produce any results. So, I took N=24 and got the result.
Maybe p, q, r, s > 1000000, but I don't think so.
Some close result for N=12
y=14
p=7
q=5
r=13
but s = 9
What is wrong with N=12 for 2nd task, I don't know.

 

[Dr.Peterson]

We use three vertical dot to mean "divided by" and I don't know how to replace it. Perhaps, it's better to just write "divided by".
And "x+1+6=2s|:2" means that all equation need to be divided on 2.

Just for your information, in America at least, we usually use "/" to mean "divided by" in ordinary typing, and "÷" at an elementary level. I am familiar with using the colon, ":", for division, but don't think I've ever heard a three-dot colon! Symbols vary more than most people are aware of.

I did figure out what you meant by "|:"; in my experience it is best never to indicate "dividing an entire equation", which usually confuses people.

The important thing here really is that whatever symbols you use, N:2 means "N divided by 2", not "N is divisible by 2", which is entirely different. Again, this could be different where you are, but I think it is an important distinction. By I understand what you are writing.

 

[Dr.Peterson]

For 12
1) x+12=3r|:3
p+4=r and It's possible
x+1+12=2s|:2
q+6=s and It's possible too
So, N>=12.
2) y+12=2r|:2
p+6=r and It's possible
y+1+12=3s|:3
q+4=s and It's possible too
So, N>=12 too.

I understand that N is a multiple of 12 (though I don't think you've really shown that it can't be an odd multiple of 6 yet). To show that it actually is possible (not just "not impossible") for N=12, you have to show the actual solution (as you will do next).

Let me find a solution for (1) manually, in order to get a better feel for the problem.

We have p + 4 = r and q + 6 = s, and all 4 numbers have to be prime. Also, since x = 3p , x+1 = 2q, we have q = (3p + 1)/2.

So we just have to go through a list of primes for p and for q, finding pairs of primes with the right differences. For p and r, we can have (3,7), (7,11), (13, 17), ...; for each of these, we have to find q, which is, respectively, q = 5, 11, 20, and only the last is not a prime. Now we can find s, which is respectively s = 11, 17, ...

Your first solution was p=3, r=7, q=5, s=11; x=9. Check: 9 = 3p; 10 = 2q; 21 = 3r; 22 = 2s.

My second is p=7, r=11, q=11, s=17; x=21. Check: 21 = 3p; 22 = 2q; 33 = 3r; 34 = 2s.

Either of these shows that N=12 is valid. And I didn't need a program. I especially didn't need to try all quartets of primes!

So N>=12, and I found such x with a program help. For this I checked all combinations of prime numbers p, q, r, s if p, q, r, s < 1000000.
But for y (2nd task) the program did not produce any results. So, I took N=24 and got the result.
Maybe p, q, r, s > 1000000, but I don't think so.
Some close result for N=12
y=14
p=7
q=5
r=13
but s = 9
What is wrong with N=12 for 2nd task, I don't know.

I suppose you are saying that this is not a test problem that you have to do without a computer, but a problem for which any tools are allowed.

The hard part of this, of course, is to find a solution where p, q, r, and s are all prime; I don't immediately know of any other techniques. But if the computer is doing all the work, then you have no evidence from which to observe what "goes wrong". Giving at least this one "near miss" is a start.

Let me try solving (2) for N=12 to see what happens.

We have p+6=r and q+4=s; all 4 numbers have to be prime. Also, since y = 2p , y+1 = 3q, we have q = (2p + 1)/3.

Again, we can take each prime for p and find r, q, and s in turn:

[FONT=courier new] p  r  q   s
 3  [U]9[/U]  -   -
 5 11 [U]11/3[/U] -
 7 13  5   [U]9[/U]
11 17 [U]23/3[/U] -
13 19  [U]9[/U]   -[/FONT]

So there are two things that go wrong here: sometimes q is not an integer, and sometimes r or q or s is composite.

The thing to ponder is, is there some reason one of these things will always happen. I wonder if 2p+1 being a multiple of 3 forces s to be composite?

Let's keep thinking.

 

[Yarem4uk]

I used your idea, Dr.Peterson, about q = (2p + 1)/3, but changed a bit.
From the equations in (2) I concluded, that q is least prime number from the q, p, r, s (y = 2p and y+1 = 3q).
When y+1 = 3q=> y = 3q-1.
Then I solved the remaining equations by q
3q-1=2p=> p=(3q-1)/2
3q-1+12=2r=> r=(3q+11)/2
3q-1+13=3s=>s=q+4
But I did not find any solution, when q is prime and s, p, r are primes too for q<100. So I decided to write a new program.
With the help of the sieve of Eratosthenes I created an array of prime numbers and called it "primes" (numbers in array are <10000).
Then I consistently took from the array the number for q, calculated p, r, s and y. If the numbers p, r, s (q is already prime) are primes (from array "primes"), then I bring them to the screen.

When numbers in array are <10000, then there is no result. I was so surprised, when I got the result for the numbers <100000! This q, p, r and s were so far away!
q= 10429.0 s= 10433.0 p= 15643.0 r= 15649.0 Y= 31286.0
q= 16069.0 s= 16073.0 p= 24103.0 r= 24109.0 Y= 48206.0
...
Thank you very much, Dr.Peterson, for all the help! I solved this task for several days!