# elementary number theory – Show \$ gcd (a, b) = 1 \$ implies \$ varphi (a cdot b) = phi (a) cdot phi (b) \$ This means that if $$gcd (m, n) = 1$$, then $$φ (mn) = φ (m) φ (n)$$, (Evidence: let $$A, B, C$$ are the sets of nonnegative integers that are respectively too and less than coprime $$m, n$$, and $$Mn$$; then there is a bijection between $$A × B$$ and $$C$$, according to the Chinese remainder theorem.)

I also saw this on the wiki page of Euler's Totient function, but I had no idea$$dots$$

My experiments:

After FTA we have:
$$a = p_1 ^ { alpha_1} cdots p_n ^ { alpha_n}$$
$$b = q_1 ^ { beta_1} cdots q_m ^ { beta_m}$$
Since $$gcd (a, b) = 1$$, to have $$p_i neq q_j$$, Where $$i in (1, n), j in (1, m)$$implies:
$$a cdot b = p_1 ^ { alpha_1} cdots p_n ^ { alpha_n} cdot q_1 ^ { beta_1} cdots q_m ^ { beta_m}$$
From that:
begin {align} & ~~~~~~ varphi (a cdot b) \ & = varphi (p_1 ^ { alpha_1} cdots p_n ^ { alpha_n} cdot q_1 ^ { beta_1} cdots q_m ^ { beta_m}) tag * {(1)} \ & = a cdot b (1 frac {1} {p_1}) cdots (1 frac {1} {p_n}) (1 frac {1} {q_1}) cdots (1 frac {1} {q_m}) tag * {(2)} \ & = a (1- frac {1} {p_1}) cdots (1- frac {1} {p_n}) cdot b (1- frac {1} {q_1}) cdots (1- frac {1} {q_m}) tag * {(3)} \ & = varphi (p_1 ^ { alpha_1} cdots p_n ^ { alpha_n}) cdot varphi (q_1 ^ { beta_1} cdots q_m ^ { beta_m}) tag * {(4)} \ & = varphi (a) cdot varphi (b) tag * {(5)} end

Is this proof valid since I have seen the proof of Euler's product formula?$$($$used on step $$(2))$$ It seems like I would use this feature too. If I then use Euler's product formula to prove this property, it seems a bit circular, or are there other approaches $$?$$ Posted on Categories Articles