# Thread: Find gcd of 7^(4n+3) − 4^(10n+3) + 11^(10n+2)

1. ## Find gcd of 7^(4n+3) − 4^(10n+3) + 11^(10n+2)

Find gcd of $\displaystyle 7^{4n+3}-4^{10n+3}+11^{10n+2}$, $\displaystyle n \in N_0$.

Any idea?

2. ... the GCD of that expression and which other one?

3. Originally Posted by coquitao
... the GCD of that expression and which other one?
I meant GCD of the results of this expresion for all n.

4. I gooot you... Then, the idea is the following: prove that your expression is always divisible by 400 = 16 x 25. Since 400 belongs to the range of it, you are done.

5. Um, just wondering how you got 400 ... cuz I got it by setting n = 0, but when n = 1, it's not divisible by 400 ... but it is divisible by 200 ...

6. Of course ... I proved the expression always to be divisible by 8 not 16. You are right, Bingk.

7. You're welcome ... but really, I just did trial and error ... how did you prove that it's always divisible by 8?

8. The second term is always divisible by 8, so the problem reduces to investigate whether $\displaystyle 7^{4n+3}+11^{10n+2}$ is divisible by 8 or not.

Since $\displaystyle 7^{2}$ is 1 modulo 8 we have that $\displaystyle 7^{4n}$ is also 1 modulo 8. Therefore $\displaystyle 7^{4n+3} \equiv 7 \equiv -1 \mod 8$.

On the other hand, the fact that $\displaystyle 11^{2} \equiv 1 \mod 8$ implies at once that $\displaystyle 11^{10n+2} \equiv 1 \mod 8$. Hence $\displaystyle 7^{4n+3}+11^{10n+2} \equiv (-1)+1 = 0 \mod 8$ as was to be shown.