This exercise has also been posted yesterday at the cryptography help thread. Since my deadline for delivery has approached earlier I decided to move it here. I don't believe it is related to cryptography as much as Number's Theory.
Prove that the following is true:
11^n + 5^n = 0 (mod7) when n = 3 (mod6)
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
Thanks in advance for your time and help.
Ya moo, he is wrong. It should actually read:
11^3 = 1 (mod7) and 5^3 = -1 (mod7)
n = 3 (mod6) means 6|(n-3). But this means 3|n. Thus n = 3k where k is an integer.
So 6|3(k-1) and thus 2|k-1 and therefore k is odd.
Now since k is odd,
11^n + 5^n = 11^3k + 5^3k = 1 + (-1)^k = 1 + -1 = 0 (mod7)
Aliter:Induction:
For n=1, follows from hint.
Assume for n and prove for n+6
Read the below line modulo 7,
11^(n+6) + 5^(n+6) = 11^n 11^6 + 5^n 5^6 = 11^n (11^3)^2 + 5^n (5^3)^2 = 11^n (1)^2 + 5^n (-1)^2 = 11^n + 5^n
But 11^n + 5^n mod 7 = 0, by induction hypothesis
Hello guys and thanks for your help
I'm sorry but the exercise as given is like this:
Prove that the following is true: 11^n + 5^n = 0 (mod7) when n = 3 (mod6)
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
I'm currently reading your posts and I see your point, the exercise has definitely some fault. Thanks for your time and effort!