
#1
Feb2312, 04:22 AM

P: 61

1. The problem statement, all variables and given/known data
Find all solutions in nonnegative integers x, y, z of the equation 2^x + 3^y= z^2 2. Relevant equations 3. The attempt at a solution not entirely sure how to go about it, im 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)) 



#2
Feb2312, 04:25 PM

Sci Advisor
HW Helper
P: 2,020

Where is this from? The first thing I tried worked, but it produced a fairly longwinded 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.




#3
Mar112, 12:42 PM

P: 61

wasn't too sure what you mean with the modulo 8,
but this is what i have so far so z cant 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^23^y has to be a perfect square then y has to be even and a power of 2 i.e. 2^x=z^23^y → 2^x = (z3^2^w)^2 



#4
Mar212, 11:12 AM

P: 61

find all solutions of the equation 2^x+3^y=z^2
so far i have
2^x + 3^y = z^2 [1] where x, y, and z are nonnegative 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+13^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? 



#5
Mar912, 12:09 AM

Sci Advisor
HW Helper
P: 2,020

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.) 



#6
Mar1212, 06:03 AM

P: 61

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 nonnegative 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)(cb)and c+b and cb must be even since c and b are odd. So therefore there must exist u,v,w such that a=2u,c+b=2v,cb=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 vw, but c+b +cb=2v+2w so 2c=2v+2w which means that c=v+w so d divides c. Also c+b(c+b)=2v2w. So 2b=2v2w which means that b=vw, 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=vw=p^2q^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^2q^2 )^2 so z=c=v+w=p^2+q^2,3^(y/2)=b=vw=p^2q^2 and 2^((x/2) )=a=2u=2pq 



#7
Mar2812, 10:18 PM

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P: 2,020

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^2m^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^(x2)  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 ##(x2,y/2) \in \{(1,0), (2,1)\}##. You can take it from here. 


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