Find all solutions of the equation 2^x+3^y=z^2

[.d9n.]

Homework Statement

Find all solutions in non-negative integers x, y, z of the equation
2^x + 3^y= z^2

Homework Equations

The Attempt at a Solution

not entirely sure how to go about it, I am assuming maybe logarithms?

2=log(base Z) (2^x+3^y)

or maybe substitution

let
u=2^x so x=log(base 2)(u)
v=3^y so y=log(base 3)(v)
w=z^2 so 2=log(base z)(w)

u+v=w → 2^(log(base 2)(u))+3^(log(base 3)(v))=2^(log(base z)(w))

 

[morphism]

Where is this from? The first thing I tried worked, but it produced a fairly long-winded solution so I'm reluctant to share it in case there's a simpler method... But here goes: Begin by considering your equation modulo 8, keeping in mind that z can't be divisible by 2.

 

[.d9n.]

wasn't too sure what you mean with the modulo 8,
but this is what i have so far
so z can't be even as its equal to an odd plus an even number = odd
so when z=0, x=3 then z=3 and when y=2, x=4 then z=5
so if z^2-3^y has to be a perfect square then y has to be even and a power of 2 i.e.
2^x=z^2-3^y → 2^x = (z-3^2^w)^2

 

[.d9n.]

so far i have

2^x + 3^y = z^2 [1]
where x, y, and z are non-negative integers.
We can see that [1] is a congruence modulo 8:
3^y = z^2 (mod 8) [2]
When z=1 and and y=2 we have
3^2 = 1 (mod 8) [3]
and by Fermat's theorem. So this means that y must be even.
As y is even, we can write y = 2k for some integer k. So we have:
3^2k + 2^x = z^2
z^2 - 3^2k = 2^x
(z + 3^k)(z - 3^k) = 2^x [4]
This means that the left hand side must be a power of 2, and therefore each factor i.e. (z+3^k) must also be a power of 2. But its not possible to have both factors on the left hand side divisible by 8, as this would imply that their difference
(z + 3^k) - (z - 3^k) = 2*3^k
would be divisible by 8, which is clearly impossible. The only power of 8 that is not a multiple of 8 is 8^0 = 1. So therefore one of the factors must be 1, and this is obviously the smaller factor; we have:
z- 3^k = 1
z = 3^k + 1 [5]
If we substitute that into [4], we obtain:
(3^k+1+3^k )(3^k+1-3^k )=2^x
2∙3^k + 1 = 2^x
2∙3^k = 2^x - 1 [6]
and this means that 2^x - 1 must be twice a power of 3.
So if k = 0, we have 2^x - 1 = 2, which doesn’t work. If k = 1, we have:
6 = 2^x - 1
where the LHS comes out even and the RHS comes out odd, so think i may have gone wrong somewhere, any ideas?

 

[morphism]

Sorry, I'm too tired to read all that, so let me indicate what I'd meant with my hint.

Looking at the equation mod 8 and mod 3, and using the fact that $2\nmid z$ and $3\nmid z$, we can conclude that x and y must be even. Thus the equation $$ (2^{x/2})^2 + (3^{y/2})^2 = z^2 $$ is primitive Pythagorean, and the solutions to such equations are fully known (google it).

If you write down a parametrization for these Pythagorean triples, and use the fact that the only integer solutions to $$2^a = 3^b + 1$$ are $(a,b) \in \{(1,0), (2,1)\}$ (easy to prove), you should be able to obtain all positive integral solutions (x,y,z) to the original equation. (I got (3,0,3) and (4,2,5) as the only solutions.)

 

[.d9n.]

this is what i have so far, but think i may have gone wrong somewhere, any ideas?We are looking for solutions of:
2^x + 3^y = z^2 [1]
where x, y, and z are non-negative integers.
We can see that [1] is a congruence modulo 8:
3^y = z^2 (mod 8) [2]
When z=1 and and y=2 we have
3^2 = 1 (mod 8) [3]
and by Fermat's theorem. So this means that y must be even.
As y is even, we can write y = 2k for some integer k. So we have:
3^2k + 2^x = z^2
z^2 - 3^2k = 2^x
(z + 3^k)(z - 3^k) = 2^x
We can see that [1] is a congruence modulo 3:
2^x = z^2 (mod 3) [4]
When z=1 and and y=2 we have
2^x = 1 (mod 3) [5]
and by Fermat's theorem. So this means that x must be even.
As x is even, we can write x = 2k for some integer l. So we have:
3^y + 2^2l = z^2
z^2 - 2^2l = 3^y
(z + 2^l)(z - 2^l) = 3^y
So we are left with
2^2l + 3^2k = z^2
(2^l )^2+ (3^k )^2=z^2 [6]
Which is a Pythagorean triple, where a^2+b^2=c^2. Where a=2^(x/2),b=3^(y/2) and c=z, we can assume that a,b,c are coprime due to Fermat’s Lemma.
We know that c has to be odd because if we assume c is even, then there exists another value C such that c=2∙C. Also that c^2 is divisible by 4 since
c^2=(2C)^2 =4C^2
We know a and b must be odd because a,b,c are coprime. As a and b are odd there must exist values A and B such that
a=2A+1,b=2B+1
but a^2+b^2 cannot be divisible by 4 since
a^2+b^2=(2A+1)^2+(2B+1)^2=4A^2+4A+1+4B^2+4B+1=4(A^2+A+B^2+B)+2
we have a contradiction so c is odd.
So we have a^2=(c+b)(c-b)and c+b and c-b must be even since c and b are odd. So therefore there must exist u,v,w such that
a=2u,c+b=2v,c-b=2w
which means that (2u)^2=(2v)(2w) dividing both sides by 4 gives us u^2=vw.
If we assume v and w are not coprime, then there exists d such that d>1 and d divides both v,w, then d divides both v+w and v-w, but
c+b +c-b=2v+2w
so 2c=2v+2w which means that c=v+w so d divides c.
Also c+b-(c+b)=2v-2w. So 2b=2v-2w which means that b=v-w, so d divides b. Which is a contradiction as c and b are coprime from [6].
So we reject our assumption and v and w are coprime. By properties of coprimes we know that v and w are squares of themselves. So there exists p,q such that
v=p^2,w=q^2
and we have our solutions since
c=v+w=p^2+q^2
b=v-w=p^2-q^2
a=2u=2pq (As u^2=vw which means u=pq)
We know p,q are relatively prime and opposite parity as z is odd, so we are left with.
(p^2+q^2)=(2pq)^2+(p^2-q^2 )^2

so z=c=v+w=p^2+q^2,3^(y/2)=b=v-w=p^2-q^2 and 2^((x/2) )=a=2u=2pq

 

[morphism]

Okay, so we know that 2^(x/2), 3^(y/2) and z are a primitive Pythagorean triple. This means that we can write 2^(x/2)=2nm, 3^(y/2)=n^2-m^2 and z=n^2+m^2 for some integers n and m, with only n even (say). From the first of these equations we find that m=1 and n=2^(x/2 - 1). But then the second equation becomes 3^(y/2) = 2^(x-2) - 1.

This is an equation of the form 2^a = 3^b + 1, and we are looking for integral solutions to such an equation. This forces $b\geq0$ (for otherwise 2^a won't be an integer), which in turn implies that $a\geq1$. I claim that $a \leq 2$ as well. Indeed, if a>2, then 3^b+1=0 mod 8, and this leads to a contradiction. Thus a is either 1 or 2. From here it's easy to see that the only possible solutions (a,b) are (1,0) and (2,1).

Returning to our problem, we see that $(x-2,y/2) \in \{(1,0), (2,1)\}$. You can take it from here.