if 2^k+2 + 3^3k is divisable by 5 then 2^k+3 + 3^3k+3 is also divisible by 5 use direct proof
Let's assume that 2^k+2 + 3^3k is divisible by 5
Therefore, 2^k+2 + 3^3k ≡ 0 (mod 5)
We want to prove that 2^k+3 + 3^3k+3 is also divisible by 5
Using the laws of exponents, we can simplify the expression as follows:
2^k+3 + 3^3k+3 = 2*2^k+2*2 + 3*3^3k*3^3
= 2(2^k+2) + 27(3^3k)
Since 2^k+2 + 3^3k is divisible by 5, we can write it as 2^k+2 + 3^3k = 5n, where n is an integer
Substituting this expression into our simplified expression, we get:
2(5n - 3^3k) + 27(3^3k) = 10n + 24(3^3k)
= 10n + 2(5*24)(3^3k)
= 10n + 240(3^3k)
= 10(n + 24*3^3k)
Since n is an integer, n + 24*3^3k is also an integer
Therefore, 2^k+3 + 3^3k+3 is also divisible by 5, which completes the proof.