A number theory problem by Dylan Pentland

Number Theory Level 5

$\large \displaystyle a{b}^{a}=b{a}^{b}$

We are given the equation above for natural numbers $a$ and $b$ (not necessarily distinct).

How many solutions for $(a,b)$ are there if $a+b\le1000$ ?

The answer is 2496.

2 solutions

Dylan Pentland
May 20, 2015

Consider $a=b+k$ for $k>0$ . $\displaystyle (b+k)\left( {b}^{b+k} \right) = (b) {\left(b+k\right)}^{b}$ At $b=1$ , there is equality and so any value of $a$ is acceptable. By symmetry, we now have the two cases $a=1$ and $b=1$ .

These two functions are both increasing at different rates after $b=1$ with $k>0$ , meaning that one is always greater than the other and there cannot be equality. However, at $k=0$ these rates are the same giving the case $a=b$ .

To start with, exclude $(a,b)=(1,1)$ . From the first case there are $499$ solutions, the second there are $998$ and the third there are $998$ by symmetry. By adding the extra case of $(a,b)=(1,1)$ , we now know there are $500+2(998)=2496$ solutions.

Moderator note:

This is not a complete solution. It doesn't explain why "For $k > 0$ , there are no solutions other than $b = 1$ ".

Well done Dylan.Something tells me that this can also be solved using inequalities don't you think?Just an intuition by the way ...

Arian Tashakkor - 6 years ago

Yes, I think so. Consider $a=b+k$ for $k>0$ . $\displaystyle (b+k)\left( {b}^{b+k} \right) = (b) {\left(b+k\right)}^{b}$ At $b=1$ , there is equality ( $k=k$ ) and so any value of $a$ is acceptable. By symmetry, we now have the two cases $a=1$ and $b=1$ .

Dylan Pentland - 6 years ago

That is not a complete solution. It doesn't explain why "For $k > 0$ , there are no solutions other than $b = 1$ ".

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – I think my reasoning was something like this: we have two smooth increasing functions $f,g$ that grow at different rates. $f$ intersects $g$ at $x=1$ , and grows faster after $x=1$ . Hence after the intersection, $f(x)>g(x)$ . $f$ initially grows slower and $f(0)=g(0)=0$ so before $x=1$ we have $g(x)>f(x)$ .

Dylan Pentland - 5 years, 6 months ago

@Dylan Pentland – That idea is not explicit, nor has that been shown.

I agree that if $f' (k) < g ' (k)$ , then there is at most one solution (over that domain). So if we can demonstrate that and quantify when exactly we have $f' < g'$ for $k > 0$ , then since $f(0) = g(0)$ , hence there is no more solutions.

Calvin Lin Staff - 5 years, 6 months ago

I'm not sure the logs are correct. I got $\ LHS = log(ab^a) = loga +a log b \ and \ RHs = logb +bloga$

Curtis Clement - 6 years ago

You are right, but it appears the expression I made is equivalent.

EDIT: I have no clue how I got that, I think I was just fiddling around while making this question and found that the graph overlapped on the graphing calculator, and didn't think about it past that. I'm replacing it with my solution I gave Arian.

Dylan Pentland - 6 years ago

Bob Kadylo
Dec 3, 2015

If you divide both sides of the equation by ab the LHS becomes b^(a-1) and the RHS becomes a^(b-1).
Now CASE (1) for equality is a=1 and b is any value from 1 to 999.
CASE (2) for equality is b=1 and a is any value from 1 to 999.
CASE (3) for equality is a=b and that can happen 500 times (1,1),(2,2),...,(500,500) since a+b<=1000.
But the (1,1) solution has been counted in 3 times so we subtract 2.
Finally we get 2*999 + 500 - 2 = 2496

Moderator note:

It is not apparent to me why the only solutions are of the form $a = 1 , b = 1$ or $a = b$ . How does $b^{a-1} = a^{b-1}$ allow us to arrive at that conclusion?

In particular, if we wanted to solve $a^b = b ^ a$ , would you miss out $\{ 2, 4 \}$ ?

It is not apparent to me why the only solutions are of the form $a = 1 , b = 1$ or $a = b$ . How does $b^{a-1} = a^{b-1}$ allow us to arrive at that conclusion?

In particular, if we wanted to solve $a^b = b ^ a$ , would you miss out $\{ 2, 4 \}$ ?

Calvin Lin Staff - 5 years, 6 months ago

If a=1 then b has exponent 0 and LHS = 1 with RHS = 1 because we'll have 1 to an exponent. If b=1 then a has exponent 0 and RHS = 1 with LHS = 1 because we'll have 1 to an exponent. If a=b both sides are equivalent.

Bob Kadylo - 5 years, 6 months ago

I agree that "There are 3 cases where we have solutions." This tells us that we have at least 2496 solutions.

You have not explained why "There are no other cases where we have solutions." In particular, how do we know that there isn't a 2497th solution? For example, if $a = 2$ , why must we have $b = 2$ or $b = 1$ ? Why can't we have another solution?

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – Because no matter what Natural Number 'a' you give me, the ONLY two values of 'b' which make LHS=RHS is either b=a or b=1. If 'b' is any other Natural Number then you will not achieve equality in the equation. By the way, your a^b=b^a equation did appear as a Brilliant question once - and I did not miss the ordered solutions {2,4} and {4,2} because I knew how special they are. Ever since I memorized the powers of 2, I keep in mind that 2^4=16 and 4^2=16 ..... in case it might be relevant. I wonder if that is the ONLY pair of Natural Numbers for which that's true. I haven't found another such pair !! I realize that is not a proof but no counter example after many efforts implies it might be.

Bob Kadylo - 5 years, 6 months ago

@Bob Kadylo – Right, so that is where your solution has a gap.

Note that the "lack of counter example" doesn't mean "the gap is filled". It suggests that "the gap can be filled", and I believe that Dylan is on a good path towards that.

Calvin Lin Staff - 5 years, 6 months ago

A number theory problem by Dylan Pentland

The answer is 2496.

2 solutions

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