If m,n=1,show that m,φn=1
Since "\\operatorname{gcd}(m, n)=1," you know from Euler-Fermat that
"m^{\\varphi(n)} \\equiv 1 \\quad(\\bmod \\ n)"
and, similarly,
"n^{\\varphi(m)} \\equiv 1 \\quad(\\bmod \\ m)"
Since "n^{\\varphi(m)} \\equiv 0(\\bmod n)," we also have
"m^{\\varphi(n)}+n^{\\varphi(m)} \\equiv 1+0 \\quad(\\bmod \\ n)"
and, similarly,
"m^{\\varphi(n)}+n^{\\varphi(m)} \\equiv 1+0 \\quad(\\bmod \\ m)"
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