Is it possible to find natural numbers a and b that satisfy 2^a-3^b = 7?

[kaliprasad]

Find natural numbers a and b such that $2^a-3^b = 7$

 

[Nono713]

Spoiler

By observation, $2^4 - 3^2 = 7$.

 

[kaliprasad]

Spoiler

By observation, $2^4 - 3^2 = 7$.

Answer is right but I want solution

 

[Nono713]

Answer is right but I want solution

This is a solution, I've shown a pair $(a, b)$ in natural numbers which satisfies your challenge. Do you mean you want it derived analytically along with a proof that there is only one (or more, I don't know) such pair(s)? :p

 

[kaliprasad]

This is a solution, I've shown a pair $(a, b)$ in natural numbers which satisfies your challenge. Do you mean you want it derived analytically along with a proof that there is only one (or more, I don't know) such pair(s)? :p

You are right I meant solve. Sorry for wrong wording

 

[kaliprasad]

Above answer is correct
the full solution is

Spoiler

a cannot be odd because if a is odd then$2^a$ mod 3 = -1 so $2^a – 3^b$ mod 3 = -1 so it cannot be 7 as 7 = 1 mod 3b cannot be odd as if b is odd $3^b$ =3 mod 8
so $2^a – 3^b = 5$ mod 8 for a >= 3if a = 1 or 2 $2^ a< 7$ so $2^a – 3^b = 7$ not possibleso a and b both are evensay a = 2x and b = 2yso $2^{2x} – 3^{2y} = 7$or $(2^x + 3^y)(2^x- 3^y) = 7$so $2^x + 3^y = 7$ and $2^x – 3^y = 1 $solving these 2 we get x = 2 and y = 1 or a= 4 and b = 2