{{begin-eqn}} {{eqn | l = \map \phi 1 \map \tau 1       | r = 1 \times 1       | c = {{EulerPhiLink|1}}, {{TauLink|1}} }} {{eqn | r = 1       | c =  }} {{eqn | r = \map \sigma 1       | c = {{SigmaLink|1}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi 3 \map \tau 3       | r = 2 \times 2       | c = {{EulerPhiLink|3}}, {{TauLink|3}} }} {{eqn | r = 4       | c =  }} {{eqn | r = \map \sigma 3       | c = {{SigmaLink|3}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {14} \map \tau {14}       | r = 6 \times 4       | c = {{EulerPhiLink|14}}, {{TauLink|14}} }} {{eqn | r = 24       | c =  }} {{eqn | r = \map \sigma {14}       | c = {{SigmaLink|14}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {42} \map \tau {42}       | r = 12 \times 8       | c = {{EulerPhiLink|42}}, {{TauLink|42}} }} {{eqn | r = 96       | c =  }} {{eqn | r = \map \sigma {42}       | c = {{SigmaLink|42}} }} {{end-eqn}} To show that there are no more such integers, we use the formulas: {{begin-eqn}} {{eqn | l = n       | m = \prod_{p_i \mathop \divides n} {p_i}^{k_i}       | ro = =       | r = \prod_{j \mathop = 1}^r {p_i}^{k_i}       | c = Prime Decomposition }} {{eqn | l = \map \phi n       | m = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}       | ro = =       | r = \prod_{i \mathop = 1}^r p_i^{k_i} \paren {1 - \frac 1 p_i}       | c = Euler Phi Function of Integer }} {{eqn | l = \map \tau n       | m = \prod_{i \mathop = 1}^r \paren {k_i + 1}       | c = Tau Function from Prime Decomposition }} {{eqn | l = \map \sigma n       | m = \prod_{i \mathop = 1}^r \frac {p_i^{k_i + 1} - 1} {p_i - 1}       | c = Sigma Function of Integer }} {{end-eqn}} Which arises from Euler Phi Function is Multiplicative, Tau Function is Multiplicative and Sigma Function is Multiplicative. We define the function $f = \phi \cdot \tau$ as follows: :$\map f {p^k} = \map \phi {p^k} \map \tau {p^k} = p^k \paren {1 - \dfrac 1 p} \paren {k + 1}$ where $p$ is a prime, and $k$ is a non-negative integer. Note that $f$ and $\dfrac f \sigma$ are multiplicative functions, by Product of Multiplicative Functions is Multiplicative. Then $\map \phi n \map \tau n = \map \sigma n$ is equivalent to: :$\displaystyle \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map \sigma {p_i^{k_i} } } = 1$ We consider the cases $k \ge 1$. Then: {{begin-eqn}} {{eqn | l = \map f {p^k}       | o = \ge       | r = \map \sigma {p^k} }} {{eqn | ll = \leadsto       | l = p^k \paren {1 - \frac 1 p} \paren {k + 1}       | o = \ge       | r = \frac {p^{k + 1} - 1} {p - 1} }} {{eqn | ll = \leadsto       | l = p^{k - 1} \paren {p - 1}^2 \paren {k + 1}       | o = \ge       | r = p^{k + 1} - 1 }} {{eqn | ll = \leadsto       | l = k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1       | o = \ge       | r = 2 \paren {k + 1} p^k }} {{end-eqn}} Suppose equality holds. If $k \ge 2$, {{begin-eqn}} {{eqn | l = k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1       | o = \ge       | r = 2 \paren {k + 1} p^k }} {{eqn | ll = \leadsto       | l = 1       | o = \equiv       | r = 0       | rr = \pmod p       | c = since $p^{k - 1}, p^k, p^{k + 1}$ are multiples of $p$ }} {{end-eqn}}  Which is a contradiction. This forces $k = 1$. The equation becomes {{begin-eqn}} {{eqn | l = p^2 + 2 + 1       | r = 4 p }} {{eqn | ll = \leadsto       | l = p^2 - 4 p + 3       | r = 0 }} {{eqn | ll = \leadsto       | l = \paren {p - 1} \paren {p - 3}       | r = 0 }} {{end-eqn}} Since $p \ne 1$, we must have $p = 3$. Indeed, $\map f 3 = \map \phi 3 \map \tau 3 = \map \sigma 3$. For the inequality case: $k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1 > 2 \paren {k + 1} p^k$, we check each prime $p$ individually. For $p = 2$: {{begin-eqn}} {{eqn | l = k \cdot 2^{k + 1} + \paren {k + 1} 2^{k - 1} + 1       | o = >       | r = 2 \paren {k + 1} 2^k }} {{eqn | ll = \leadsto       | l = 4 k \paren {2^{k - 1} } + \paren {k + 1} \paren {2^{k - 1} }       | o = \ge       | r = 4 \paren {k + 1} \paren {2^{k - 1} }       | c = Since both sides are integers }} {{eqn | ll = \leadsto       | l = 4 k + \paren {k + 1}       | o = \ge       | r = 4 \paren {k + 1} }} {{eqn | ll = \leadsto       | l = k       | o = \ge       | r = 3 }} {{end-eqn}} For $p \ge 3$, we consider the criterion $k p \ge 2 \paren {k + 1}$. This criterion implies $k p^{k + 1} \ge 2 \paren {k + 1} p^k$, which in turn implies our inequality $k p^{k + 1} + \paren {k + 1} + 1 > 2 \paren {k + 1} p^k$. For $p = 3$ and $k \ge 2$, $3 k = 2 k + k \ge 2 k + 2$. For $p \ge 5$, $k p \ge 5 k > 2 k + 2 k \ge 2 k + 2$. Therefore $\map f {p^k} > \map \sigma {p^k}$, except for $\map f 3 = \map \sigma 3$, $\map f 2 < \map \sigma 2$ and $\map f 4 < \map \sigma 4$. Now we try to find integers $n$ satisfying $\displaystyle \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map \sigma {p_i^{k_i} } } = 1$. First observe that the sequence $\sequence {\dfrac {\map f {p^k} } {\map \sigma {p^k} } }_{k \mathop = 1}^\infty$ is strictly increasing: {{begin-eqn}} {{eqn | l = \frac {\map f {p^{k + 1} } } {\map \sigma {p^{k + 1} } } - \frac {\map f {p^k} } {\map \sigma {p^k} }       | r = \frac {p^k \paren {1 - 1 / p} \paren {k + 2} } {\paren {p^{k + 2} - 1} / \paren {p - 1} } - \frac {p^{k - 1} \paren {1 - 1 / p} \paren {k + 1} } {\paren {p^{k + 1} - 1} / \paren {p - 1} } }} {{eqn | r = \frac {p^{k - 1} \paren {p - 1}^2 \paren {k + 2} } {p^{k + 2} - 1} - \frac {p^{k - 2} \paren {p - 1}^2 \paren {k + 1} } {p^{k + 1} - 1} }} {{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p \paren {p^{k + 1} - 1} \paren {k + 2} - \paren {p^{k + 2} - 1} \paren {k + 1} } }} {{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p^{k + 2} - \paren {k + 2} p + k + 1} }} {{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {1 + \paren {k + 2} \paren {p - 1} + \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m - \paren {k + 2} p + k + 1}        | c = Binomial Theorem }} {{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m }} {{eqn | o = >       | r = 0       | c = The sum is non-empty and each term is strictly positive }} {{end-eqn}} Now we consider these particular values for $\dfrac f \sigma$: :$\dfrac {\map f 2 } {\map \sigma 2 } = \dfrac 2 3$ :$\dfrac {\map f 4 } {\map \sigma 4 } = \dfrac 6 7$ :$\dfrac {\map f 3 } {\map \sigma 3 } = 1$ :$\dfrac {\map f 5 } {\map \sigma 5 } = \dfrac 4 3$ :$\dfrac {\map f {25} } {\map \sigma {25} } = \dfrac {60} {31} > \dfrac 3 2$ :$\dfrac {\map f 7 } {\map \sigma 7 } = \dfrac 3 2$ And for $p \ge 11$: {{begin-eqn}} {{eqn | l = \frac {\map f p } {\map \sigma p }       | r = \frac {2 \paren {p - 1} } {\paren {p^2 - 1} / \paren {p - 1} } }} {{eqn | r = \frac {2 \paren {p - 1} } {p + 1} }} {{eqn | r = 2 - \frac 4 {p + 1} }} {{eqn | o = \ge       | r = 2 - \frac 4 {12} > \frac 3 2 }} {{end-eqn}} By the strictly increasing property above, every higher power of these primes has $\dfrac f \sigma > \dfrac 3 2$. Since $\dfrac f \sigma$ is multiplicative, only $n = 3, 14, 42$ can give $\dfrac {\map f n} {\map \sigma n} = 1$. Since $\map \phi 1 \map \tau 1 = \map \sigma 1$, the only positive integers whose Euler $\phi$ function multiplied by its  $\tau$ function equals its $\sigma$ function are $1, 3, 14, 42$. {{qed}}