positive integers

tony123 · 2009-06-22 10:24:14

Determine all pair(s) (A, B) of positive integers that satisfy this equation.

Last edited by tony123 (2009-06-22 10:26:20)

mathsyperson · 2009-06-22 11:00:18

I'm pretty sure (1,1) is the only solution.

bobbym · 2009-06-22 12:06:34

You are right and I think we can prove it like this.

If A = B and A,B are positive integers then it is clear the only solution to

is 1

When A ≠ B and A,B are positive integers then we assume that

and work to a contradiction.

Take the log of both sides.

Divide both sides by log(A)

Now since is an integer and could never be an integer.

We have a contradiction. So there are no other solutions besides (1,1).

Last edited by bobbym (2009-06-23 06:39:03)

wintersolstice · 2009-06-22 20:36:18

bobbym wrote:

Divide both sides by log(A)

Now since
is an integer and
could never be an integer.
We have a contradiction. So there are no other solutions besides (1,1).

If B is a power of A then it

can be an integer so that's not a contradiction.

Last edited by wintersolstice (2009-06-22 20:37:08)

wintersolstice · 2009-06-22 21:09:12

bobbym wrote:

You are right and I think we can prove it like this.

I think I can prove that A must equal B by extending your proof (your prove was just incomplete) and I know of another solution (0,0)!

which means that:

Last edited by wintersolstice (2009-06-22 21:12:07)

wintersolstice · 2009-06-22 22:38:28

I hope I'm not breaking any rules here (two many consecutive posts!) I just spotted a mistake in my proof that makes me think there are other solutions!

mathsyperson · 2009-06-22 23:32:21

0^0 is a tricky creature. Some people insist that it's equal to 1, some think it's 0 and others say it doesn't have a value at all.

Without getting into that debate though, (0,0) is not a solution because tony's interested in positives.

wintersolstice · 2009-06-22 23:34:58

mathsyperson wrote:

0^0 is a tricky creature. Some people insist that it's equal to 1, some think it's 0 and others say it doesn't have a value at all.
Without getting into that debate though, (0,0) is not a solution because tony's interested in positives.

saying that 0^0 = 1 is actually incorrect (I read it somewhere!)

plus there's (-1,-1)

also there's something very completed that I discovered that could uncover another solution

if 0^0 = 1 then 0/0 = 1

this means that
(0/0)*5 = (0/0) would simplify to 5=1 (this is a paradox)

Last edited by wintersolstice (2009-06-22 23:37:57)

bobbym · 2009-06-22 23:44:17

Hi wintersolstice;

You are right, when B is a power of A then it is an integer. There is no contradiction there. Thanks for pointing that mistake out. I still don't think there aren't any other solutions other than (1,1) among the positive integers. Post if you find any other solutions.

Last edited by bobbym (2009-06-22 23:45:38)

wintersolstice · 2009-06-23 00:10:40

mathsyperson wrote:

Without getting into that debate though, (0,0) is not a solution because tony's interested in positives.

Ok I didn't read the puzzle properly and sorry for going on about the "0^0" debate

I think it's safe to assume that (1,1) is the only solution for A=B I think I have something that could help see if there's solutions for A≠B. (but it's complicated)

Should I explain my idea?:D

wintersolstice · 2009-06-23 00:21:46

In fact the method I've got just re-writes the puzzle so it's pointless.

bobbym · 2009-06-23 01:03:13

So, we still need some demonstration that (1,1) is the only answer.

TheDude · 2009-07-07 04:43:40

Reviving this problem, you can carry bobby's work a little further. For the sake of redundancy I want to point out that both A and B must be greater than 1 for logs to be meaningful, otherwise you get into divide by 0 problems. Not a huge issue since you can test A = B = 1 by hand, just wanted to cover all of our bases.

Now, we got to

The left side is clearly an integer. For the right side to also be an integer we must have B = A^n, where n is a rational number greater than 1 (I won't show the work here, but you can pretty easily see that B > A and thus n > 1). Rewriting that equation gives us

Now take the log again:

There are 3 possibilities: n < A, n = A, and n > A. If n = A then the LHS is 0 but the RHS is 1, so n cannot be equal to A. If n > A the LHS is negative but the RHS is positive, so n cannot be greater than A. If n < A the RHS is less than 1, so we can narrow our search to A - 1 < n < A.

Off the top of my head I can't think of a way to prove that the solution is unsolvable, but I'm pretty sure it is.

Naveen Ks alias fraggy · 2009-09-23 05:26:55

A^[A^A]=B^B, since all are intgrs
A^A=A this i guess shoulld be true,[dis happens when u dont rearrange]
it only happens whenA=1;

Math Is Fun Forum

#1 2009-06-22 10:24:14

positive integers

#2 2009-06-22 11:00:18

Re: positive integers

#3 2009-06-22 12:06:34

Re: positive integers

#4 2009-06-22 20:36:18

Re: positive integers

#5 2009-06-22 21:09:12

Re: positive integers

#6 2009-06-22 22:38:28

Re: positive integers

#7 2009-06-22 23:32:21

Re: positive integers

#8 2009-06-22 23:34:58

Re: positive integers

#9 2009-06-22 23:44:17

Re: positive integers

#10 2009-06-23 00:10:40

Re: positive integers

#11 2009-06-23 00:21:46

Re: positive integers

#12 2009-06-23 01:03:13

Re: positive integers

#13 2009-07-07 04:43:40

Re: positive integers

#14 2009-09-23 05:26:55

Re: positive integers

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