# The famous 2^4 = 4^2 problem

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• October 28th 2009, 07:45 PM
Some1Godlier
The famous 2^4 = 4^2 problem
2^4 = 4^2 (X^Y=Y^X) I have to prove that (2,4) and (4,2) are the only pairs of distinct positive whole numbers having that property. I would also like to change the function into parametric but am quite lost. I have found some information concerning this topic but as of now it seems like gibberish (Can anyone help explain??):

First take the log of both sides:

log(X^Y) = log(Y^X)

and simplify:

Y*log(X) = X*log(Y)

and then divide by X*Y:

log(X) log(Y)
------ = ------.
X Y

Now you should consider the function

log(x)
f(x) = ------.
x

Clearly, we have a solution to the last equation if and only if

f(X) = f(Y).

Well, this happens when X = Y, but does it happen elsewhere? If we graph

y = f(x)

we will find that f increases from y = -infinity at x = 0 to y = 1/e
at x = e (that's e = 2.71828... whether you used the common log or the
natural log or the log to any other base), and then f decreases from
y = 1/e at x = e to y = 0 at x = infinity.

Well, if X and Y are different values and

f(X) = f(Y),

then that means that there is a horizontal line which passes through
our function at two points (namely X and Y). Look at the function,
and you'll find that the smaller value is somewhere between 1 and e,
and the larger value is bigger than e. Also, the closer the smaller
value is to e, the closer the larger value is to e. The closer the
smaller value is to 1, the bigger the larger value is.

So what you find is that if X <= 1, then the only solution is Y = X.
Similarly, if X = e, then the only solution is Y = X. But if

1 < X < e,

then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number bigger than e. Similarly, if

e < X,

then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number between 1 and e.

But can you write out a formula for the smaller value in terms of the
bigger value, or vice-versa? Well, not using any closed-form
function. But you can use numerical methods to find approximate
solutions for any X value.

you might be interested in. The solutions you gave (2, 4) and (4, 2)
are in integers. In fact, these solutions and X = Y are the only
solutions in positive integers. And the only integer solutions are X
= Y and (2, 4), (4, 2), (-2,-4), (-4, -2).

Proving that is as follows: First suppose that X and Y are positive.
By switching the order of X and Y, we may assume that Y >= X. Now
divide both sides of the equation by X^X.

X^(Y-X) = (Y/X)^X.

Since the left side is clearly an integer, the right side has to be an
integer. But if you raise a non-integer rational number to an integer
power, then you don't get an integer. So that means that

k = Y/X

must be an integer (bigger than 0). Now we re-write our equation as

X^(kX - X) = k^X.

We take the positive real X'th root of both sides of the equation and get

X^(k-1) = k.

Now if X >= 2, then:

(a) k = 1 always works (and means X = Y)
(b) k = 2 implies X = X^(2-1) = 2 (and gives your solutions)
(c) k = 3 implies X^(k-1) > k
(d) by induction on k, k >= 3 implies
X^(k-1) = X*X^(k-2) > X(k-1) >= 2k-2 > k,

so there are only the solutions already mentioned when X >= 2.

But X = 1 implies Y = 1. And X = 0 implies Y = 0. And if X is
negative, but Y is positive, then Y^X is positive, so X^Y is positive,
which means that Y is even and

X^Y = (-X)^Y = Y^X, so (-X)^Y * Y^(-X) = 1,

but then (-X)^Y and Y^(-X) both have to be 1, so X is -1 and Y = 1.

Finally, if X and Y are both negative, then we raise both sides to the
-1 power and get

X^(-Y) = Y^(-X)

and then if X is odd and Y is even or vice-versa, then the signs don't
match, but if X and Y are both odd, then we multiply both sides of the
equation by -1 to get

(-X)^(-Y) = (-Y)^(-X).

If both X and Y are even, then we don't need to multiply, and we still
get the same equation. So (-X, -Y) is a solution in positive integers.
• October 28th 2009, 08:06 PM
tonio
Quote:

Originally Posted by Some1Godlier
2^4 = 4^2 (X^Y=Y^X) I have to prove that (2,4) and (4,2) are the only pairs of distinct positive whole numbers having that property. I would also like to change the function into parametric but am quite lost. I have found some information concerning this topic but as of now it seems like gibberish (Can anyone help explain??):

First take the log of both sides:

log(X^Y) = log(Y^X)

and simplify:

Y*log(X) = X*log(Y)

and then divide by X*Y:

log(X) log(Y)
------ = ------.
X Y

Now you should consider the function

log(x)
f(x) = ------.
x

Clearly, we have a solution to the last equation if and only if

f(X) = f(Y).

Well, this happens when X = Y, but does it happen elsewhere? If we graph

y = f(x)

we will find that f increases from y = -infinity at x = 0 to y = 1/e
at x = e (that's e = 2.71828... whether you used the common log or the
natural log or the log to any other base), and then f decreases from
y = 1/e at x = e to y = 0 at x = infinity.

Well, if X and Y are different values and

f(X) = f(Y),

then that means that there is a horizontal line which passes through
our function at two points (namely X and Y). Look at the function,
and you'll find that the smaller value is somewhere between 1 and e,
and the larger value is bigger than e. Also, the closer the smaller
value is to e, the closer the larger value is to e. The closer the
smaller value is to 1, the bigger the larger value is.

So what you find is that if X <= 1, then the only solution is Y = X.
Similarly, if X = e, then the only solution is Y = X. But if

1 < X < e,

then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number bigger than e. Similarly, if

e < X,

then there are exactly two solutions for Y, one of which is Y = X, and
the other is some number between 1 and e.

But can you write out a formula for the smaller value in terms of the
bigger value, or vice-versa? Well, not using any closed-form
function. But you can use numerical methods to find approximate
solutions for any X value.

you might be interested in. The solutions you gave (2, 4) and (4, 2)
are in integers. In fact, these solutions and X = Y are the only
solutions in positive integers. And the only integer solutions are X
= Y and (2, 4), (4, 2), (-2,-4), (-4, -2).

Proving that is as follows: First suppose that X and Y are positive.
By switching the order of X and Y, we may assume that Y >= X. Now
divide both sides of the equation by X^X.

X^(Y-X) = (Y/X)^X.

Since the left side is clearly an integer, the right side has to be an
integer. But if you raise a non-integer rational number to an integer
power, then you don't get an integer. So that means that

k = Y/X

must be an integer (bigger than 0). Now we re-write our equation as

X^(kX - X) = k^X.

We take the positive real X'th root of both sides of the equation and get

X^(k-1) = k.

Now if X >= 2, then:

(a) k = 1 always works (and means X = Y)
(b) k = 2 implies X = X^(2-1) = 2 (and gives your solutions)
(c) k = 3 implies X^(k-1) > k
(d) by induction on k, k >= 3 implies
X^(k-1) = X*X^(k-2) > X(k-1) >= 2k-2 > k,

so there are only the solutions already mentioned when X >= 2.

But X = 1 implies Y = 1. And X = 0 implies Y = 0. And if X is
negative, but Y is positive, then Y^X is positive, so X^Y is positive,
which means that Y is even and

X^Y = (-X)^Y = Y^X, so (-X)^Y * Y^(-X) = 1,

but then (-X)^Y and Y^(-X) both have to be 1, so X is -1 and Y = 1.

Finally, if X and Y are both negative, then we raise both sides to the
-1 power and get

X^(-Y) = Y^(-X)

and then if X is odd and Y is even or vice-versa, then the signs don't
match, but if X and Y are both odd, then we multiply both sides of the
equation by -1 to get

(-X)^(-Y) = (-Y)^(-X).

If both X and Y are even, then we don't need to multiply, and we still
get the same equation. So (-X, -Y) is a solution in positive integers.

Wow! That was a huge message, and without LaTex I doubt many people will be willing to read it all.
Anway, you reached already the conclusion that it'd be wise to study the function $f(x)=\frac{\ln x}{x}$...and indeed, it is wise: check that this function has a maximum at $x=e$ and thus it gets twice all the values in the interval $(f(1),f(e))$ when $x\rightarrow \infty$
Now check that the only integers s.t. $f(x)=f(y)$ are 2,4.

Tonio
• October 28th 2009, 08:10 PM
mr fantastic
Of related interest .....
A thread of related interest: http://www.mathhelpforum.com/math-he...stion-3-a.html
• October 28th 2009, 08:24 PM
Some1Godlier
Quote:

Originally Posted by tonio
check that this function has a maximum at $x=e$ and thus it gets twice all the values in the interval $(f(1),f(e))$ when $x\rightarrow \infty$
Now check that the only integers s.t. $f(x)=f(y)$ are 2,4.

Tonio

I am quite confused. What do you mean by twice the values? Can you go just a little more in depth. I'm slowly grasping but I need a bigger push. Thanks for the help
• October 28th 2009, 08:26 PM
Some1Godlier
Also, what is meant by F(x) = F(Y)?? What is meant by F(y)?? Is that just Ln(y)/y??
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