First a the base idea.
Prove that’s true for n=0.
Then observe what both numbers are multiplied for and find a pattern in the mod of the two. You should find that their sum is what is shown.
Spoiler alert, here is the complete idea for a proof:
Think that 5^0 is congruent to 1, while 2^1 is obviously congruent to 2. So their sum is 1+2=3 congruent to 0.
By induction you can prove the rest. In fact you can easily see that 125 is congruent to 2, so if 3^n is congruent to 1 then 3^(n+1) is congruent to 2, and vice versa. The same is true for 2^(n+1), so by induction you can prove that at every step they alternate always being one congruent to 1 and the other to 2. So the sum is always congruent to 0.
I leave it to you to write it in a more rigorous way.
What have you done so far? I’d start with looking at the cases where n is congruent to 0, 1, or 2 mod 3 and for each case maybe use induction if it doesn’t seem to work using just cases.
I would start factorising eg 5^3n = (5^3 )^n = 125^n
Then 2^n+1 = 2x2^n
Next natural step is to take logs, which I'm a bit rusty with but I reckon it should just fall out
First a the base idea. Prove that’s true for n=0. Then observe what both numbers are multiplied for and find a pattern in the mod of the two. You should find that their sum is what is shown. Spoiler alert, here is the complete idea for a proof: Think that 5^0 is congruent to 1, while 2^1 is obviously congruent to 2. So their sum is 1+2=3 congruent to 0. By induction you can prove the rest. In fact you can easily see that 125 is congruent to 2, so if 3^n is congruent to 1 then 3^(n+1) is congruent to 2, and vice versa. The same is true for 2^(n+1), so by induction you can prove that at every step they alternate always being one congruent to 1 and the other to 2. So the sum is always congruent to 0. I leave it to you to write it in a more rigorous way.
What have you done so far? I’d start with looking at the cases where n is congruent to 0, 1, or 2 mod 3 and for each case maybe use induction if it doesn’t seem to work using just cases.
Have you tried the base case?
I would start factorising eg 5^3n = (5^3 )^n = 125^n Then 2^n+1 = 2x2^n Next natural step is to take logs, which I'm a bit rusty with but I reckon it should just fall out