True or False (1)

Let m be an integer such that $\gcd(m,6) = 1.$ Note that this also implies $\gcd(m, 2) = 1 \text{ and } \gcd(m, 3) = 1.$

Let $n := \phi(m)-1.$

I claim that $2n +3n +6n ≡ 1 (mod 15)$ for this choice of n (solving the problem).

Since $\gcd(m, 6) = 1$ , it suffices to check this after multiplying both sides by 6, i.e. we want to show that

$6·2^n+6·3^n+6·6^n≡6 (mod15).$

We have $6·2^n+6·3^n+6·6^n = 3·2^{\phi(m)}+ 2·3^{\phi(m)}+ 6^{\phi(m)}$

and by Euler’s theorem this is equivalent to $3 + 2 + 1 = 6 (mod \ m)$ , as desired.