1sorry thats wrong. see this,
suppose y≥2 , taking mod taking mod 4 gives x is odd and taking mod 3 gives y is even.
note that 3(x+12)2 = 3(2y)mod7 but (3/7)(2/7)^y =-1 is a quadratic residue mod7 but 3 is not . hence x=2 y=1.
Find all positive integers x and y satisfying the equation
3^x-2^y=7
-
UP 0 DOWN 0 0 8
8 Answers
71Don't know about others but trivial one is x = 2, y = 1
17+2y = 3x
taking mod 9 (for x ≥ 2) we can see that (2,1) is the only solution.
1
1do yo agree with me that if y≥2 then x is 2k-1?
is so then, in mod 7 we have 32k ≡ 3.2y
in mod 7 we have 2≡32 so 32k ≡ to 3.32y so 3≡ (3k3y)2
but 3 is non quadratic residue mod7 i.e. 37 = -1 , it means that there is no a such that 3≡a2. so we arrive at desired contradiction when y>1
1oh i forgot to ask, i hope you know about Legendre symbols.
and theres a typo in my post #5 i meant y is also odd.
11If y≥3 then 2y+7≡7(mod8) But 3x≡1,3(mod8), thus inconsistent, so y≤2...Verifying with y=1,2 we arrive at the unique soln (2,1)