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### One Proof that a Diophantine Equation in Two Variables Has Only Three Solutions

```Date: 01/30/2011 at 03:17:22
From: Sandesh
Subject: Diophantine Equation with exponents

Find all integers x and y for which (2^x) + (3^y) is a perfect square.

I found some solutions, but I want a full method to solve it
comprehensively.

```

```
Date: 01/30/2011 at 23:36:45
From: Doctor Vogler
Subject: Re: Diophantine Equation with exponents

Hi Sandesh,

Thanks for writing to Dr. Math. That's not an easy problem for a
14-year-old!

To prove that there are only three solutions, you must first familiarize
yourself with modular arithmetic ...

Mod, Modulus, Modular Arithmetic
http://mathforum.org/library/drmath/view/62930.html

... as well as Pythagorean triples

Pythagorean triples
http://mathforum.org/dr.math/faq/faq.pythag.triples.html

Here is how I would prove it:

We want to find integer solutions to the equation

(2^x) + (3^y) = m^2.

First of all, if x or y is negative, then (2^x) + (3^y) won't be an
integer. Next, if x = 0, then

3^y = m^2 - 1 = (m - 1)(m + 1).

So both m - 1 and m + 1 are powers of 3. Since their difference is 2, the
only possibility is

m - 1 = 1

and

m + 1 = 3.

So

m = 2
x = 0
y = 1

Next, if y = 0, then

2^x = m^2 - 1 = (m - 1)(m + 1).

So both m - 1 and m + 1 are powers of 2. Since their difference is 2, the
only possibility is

m - 1 = 2

and

m + 1 = 4

So

m = 3
x = 3
y = 0

Now, if y > 0, then 3^y is divisible by 3, so

2^x = m^2 mod 3.

But no squares are 2 mod 3, so that means that x must be even.

If x > 0, then x >= 2 since it is even, and therefore 2^x is divisible by
4. So

3^y = m^2 mod 4.

But no squares are 3 mod 4, so that means that y must also be even.

That means that 2^x + 3^y = m^2 gives you a Pythagorean triple:

(2^(x/2), 3^(y/2), m).

Obviously, 3^(y/2) is odd, so that means that there are coprime integers r
and s such that

2^(x/2) = 2rs,
3^(y/2) = r^2 - s^2, and
m = r^2 + s^2.

But 2^(x/2) = 2rs means that both r and s must be powers of 2. And since
they are coprime, one of them must be 1. But 3^(y/2) > 0, so it must be
that

s = 1
r = 2^(x/2 - 1).

And we conclude that

3^(y/2) = r^2 - s^2
= 2^(x - 2) - 1.

Now, 2^(x - 2) is still a square, since x - 2 is even, so we can factor
the right side as

3^(y/2) = [2^(x/2 - 1) - 1][2^(x/2 - 1) + 1]

And since the product is a power of 3, that means that each term must be a
power of 3:

2^(x/2 - 1) - 1 = 3^a
2^(x/2 - 1) + 1 = 3^b

But their difference is exactly 2, so the only possibility is that the
smaller one is 1 and the larger one is 3:

2^(x/2 - 1) - 1 = 1
2^(x/2 - 1) + 1 = 3

This means that

2^(x/2 - 1) = 2.

So

x/2 - 1 = 1
x/2 = 2
x = 4

And y = 2.

So there are only three solutions, namely

(x, y) = (0, 1),
(3, 0), and
(4, 2).

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 01/31/2011 at 07:41:55
From: Sandesh
Subject: Diophantine Equation with exponents

Thank you for a very good rigorous answer.

But can you please also tell me the proof of the Pythagorean triples that
are of this form?

(2rs, r^2 - s^2, r^2 + s^2)

Also, give a proof that r and s are coprime.

```

```
Date: 01/31/2011 at 15:58:43
From: Doctor Vogler
Subject: Re: Diophantine Equation with exponents

Hi Sandesh,

Those can be found at the second link I mentioned earlier:

http://mathforum.org/dr.math/faq/faq.pythag.triples.html

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory

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