If the difference was 211, a prime number, for k =5, how would the problem be solved ? For a difference of 665, factors, 5, 7 and 19, for k =6, the methodology should work.
1) Why don't you try for x+y=65, x-y=1(actually no integer solution.) 2) You must show k is even. otherwise x=3^(k/2) and y=2^(k/2) are not integers. [Proof] 3^k - 2^k = 65 => 0 - 2^k ≡ 2 (mod 3) => 2^k ≡ 1 (mod 3) => k is even.
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For a difference of 2059, factors 71 and 29, for k=7, the solution is a matter of substitution.
Fantastic sir 🎉🎉🎉🎉🎉
K65/3/=31.265/2=31 .3 {31.2+31.3} = 62.5 2^31.5^1 2^31^1.1^1 2^1^1 2^1 (K ➖ 2k+1)
81－16=65
k=4
If the difference was 211, a prime number, for k =5, how would the problem be solved ?
For a difference of 665, factors, 5, 7 and 19, for k =6, the methodology should work.
1) Why don't you try for x+y=65, x-y=1(actually no integer solution.)
2) You must show k is even. otherwise x=3^(k/2) and y=2^(k/2) are not integers.
[Proof] 3^k - 2^k = 65 => 0 - 2^k ≡ 2 (mod 3) => 2^k ≡ 1 (mod 3) => k is even.