[proofplan] Write $N - 1 = 2^s t$ with $t$ odd and factor $N = \prod_i p_i^{e_i}$ with distinct odd primes $p_i$. For each $i$, write $p_i - 1 = 2^{a_i} v_i$ with $v_i$ odd, and let $\alpha = \min_i a_i$. The set $B$ of bases passing the strong test is shown to lie inside a subgroup of $(\mathbb{Z}/N\mathbb{Z})^\times$ whose index we compute: it is the subgroup cut out by the condition that $b$ satisfies either $b^t \equiv 1$ or $b^{2^r t} \equiv -1$ for some $r < s$, and a prime-by-prime analysis via the Chinese Remainder Theorem identifies $|B|$ as a specific product depending on the $v_i$ and $a_i$. Bounding this product shows $|B| \leq \varphi(N)/4$ whenever $N$ is composite, with the $1/4$ appearing because the worst case — $N = p q$ with two prime factors — already forces index $\geq 4$. [/proofplan]

[step:Set up the group-theoretic framework] Let $N$ be an odd composite integer. Write \begin{align*} N - 1 &= 2^s t, & N &= \prod_{i=1}^k p_i^{e_i}, \end{align*} with $t$ odd, $s \geq 1$, and $p_1, \ldots, p_k$ distinct odd primes. For each $i$ write $p_i - 1 = 2^{a_i} v_i$ with $v_i$ odd. By the [Chinese Remainder Theorem](/theorems/???) (applied to the pairwise coprime moduli $p_i^{e_i}$), \begin{align*} (\mathbb{Z}/N\mathbb{Z})^\times &\cong \prod_{i=1}^k (\mathbb{Z}/p_i^{e_i}\mathbb{Z})^\times, \end{align*} and each factor $(\mathbb{Z}/p_i^{e_i}\mathbb{Z})^\times$ is cyclic of order $\varphi(p_i^{e_i}) = p_i^{e_i - 1}(p_i - 1) = p_i^{e_i - 1} \cdot 2^{a_i} v_i$ (see [$(\mathbb{Z}/p^n\mathbb{Z})^\times$ is Cyclic for Odd $p$](/theorems/???)). Let \begin{align*} B &:= \left\{ b \in (\mathbb{Z}/N\mathbb{Z})^\times : N \text{ passes the strong test to base } b \right\}. \end{align*} Our goal is $|B| \leq \frac{1}{4} \varphi(N)$. [/step]

[step:Translate the strong-test condition into a condition on each local factor] By definition, $b \in B$ if and only if \begin{align*} b^t &\equiv 1 \pmod{N}, \qquad \text{or} \qquad b^{2^r t} \equiv -1 \pmod{N} \text{ for some } 0 \leq r < s. \end{align*} Under the isomorphism of Step 1, write $b = (b_1, \ldots, b_k)$ with $b_i \in (\mathbb{Z}/p_i^{e_i}\mathbb{Z})^\times$. The condition $b^t \equiv 1 \pmod N$ is equivalent to $b_i^t \equiv 1 \pmod{p_i^{e_i}}$ for all $i$. The condition $b^{2^r t} \equiv -1 \pmod N$ is equivalent to $b_i^{2^r t} \equiv -1 \pmod{p_i^{e_i}}$ for all $i$ — the sign $-1$ must hold locally at every prime simultaneously, because a tuple equals $-1$ in a product of rings iff it equals $-1$ in each factor. Hence \begin{align*} B &= \bigcup_{r = -1}^{s-1} B_r, \qquad B_r := \left\{ b : b_i^{2^r t} \equiv -1 \pmod{p_i^{e_i}} \text{ for all } i \right\} \quad (r \geq 0), \\ B_{-1} &:= \left\{ b : b_i^{t} \equiv 1 \pmod{p_i^{e_i}} \text{ for all } i \right\}, \end{align*} where the union is disjoint: if $b \in B_r$ and $b \in B_{r'}$ with $r < r'$, then squaring $b^{2^r t} \equiv -1$ gives $b^{2^{r+1} t} \equiv 1$, forcing $b^{2^{r'} t} \equiv 1$; but $b^{2^{r'} t} \equiv -1$, a contradiction since $N$ is odd. [/step]

[step:Count the elements of each $B_r$ via cyclic subgroup structure] Fix a local factor $(\mathbb{Z}/p_i^{e_i}\mathbb{Z})^\times$, cyclic of order $p_i^{e_i - 1} \cdot 2^{a_i} v_i$. Let $g_i$ be a generator; every element is $g_i^j$ for some $0 \leq j < p_i^{e_i - 1} \cdot 2^{a_i} v_i$. **Count for $B_{-1}$ (locally).** The condition $b_i^t \equiv 1$ is equivalent to the order of $b_i$ dividing $t$. Since $t$ is odd, this is equivalent to the order of $b_i$ dividing $\gcd(t, \varphi(p_i^{e_i})) = \gcd(t, p_i^{e_i - 1} v_i)$ (the factor $2^{a_i}$ drops because $t$ is odd). Call this common value $t_i := \gcd(t, p_i^{e_i - 1} v_i)$. The number of $b_i$ with order dividing $t_i$ in a cyclic group of order $\varphi(p_i^{e_i})$ is $t_i$. **Count for $B_r$, $r \geq 0$ (locally).** The condition $b_i^{2^r t} \equiv -1 \pmod{p_i^{e_i}}$ means $b_i^{2^{r+1} t} \equiv 1$ and $b_i^{2^r t} \not\equiv 1$. The number of $b_i$ satisfying $b_i^{2^{r+1} t} \equiv 1$ is $\gcd(2^{r+1} t, \varphi(p_i^{e_i})) = 2^{\min(r+1, a_i)} t_i$, and the number with $b_i^{2^r t} \equiv 1$ is $2^{\min(r, a_i)} t_i$. The difference counts those satisfying the exact-$-1$ condition: - if $r + 1 \leq a_i$: $2^{r+1} t_i - 2^r t_i = 2^r t_i$; - if $r + 1 > a_i$, i.e. $r \geq a_i$: the sets coincide, giving $0$. So the local count at prime $p_i$ is \begin{align*} c_{i,r} &:= \begin{cases} 2^r t_i & \text{if } r < a_i, \\ 0 & \text{if } r \geq a_i. \end{cases} \end{align*} **Global counts.** Because the conditions factor through the Chinese Remainder Theorem decomposition, \begin{align*} |B_{-1}| &= \prod_{i=1}^k t_i, \qquad |B_r| = \prod_{i=1}^k c_{i,r} \quad (r \geq 0). \end{align*} In particular $|B_r| = 0$ unless $r < \min_i a_i =: \alpha$, and for $0 \leq r < \alpha$, \begin{align*} |B_r| &= \prod_{i=1}^k 2^r t_i = 2^{rk} \prod_{i=1}^k t_i. \end{align*} [/step]

[step:Sum the counts and bound $|B|$]Summing over $r$, \begin{align*} |B| &= |B_{-1}| + \sum_{r=0}^{\alpha - 1} |B_r| = \prod_{i=1}^k t_i \cdot \left( 1 + \sum_{r=0}^{\alpha - 1} 2^{rk} \right) = \prod_{i=1}^k t_i \cdot \left( 1 + \frac{2^{\alpha k} - 1}{2^k - 1} \right). \end{align*} Simplifying the bracket, \begin{align*} 1 + \frac{2^{\alpha k} - 1}{2^k - 1} &= \frac{2^k - 1 + 2^{\alpha k} - 1}{2^k - 1} = \frac{2^{\alpha k} + 2^k - 2}{2^k - 1}. \end{align*} Meanwhile, \begin{align*} \varphi(N) &= \prod_{i=1}^k p_i^{e_i - 1} \cdot 2^{a_i} v_i \geq \prod_{i=1}^k 2^{a_i} v_i \geq 2^{k \alpha} \prod_{i=1}^k v_i, \end{align*} and $\prod_i t_i \leq \prod_i v_i$ since $t_i = \gcd(t, p_i^{e_i - 1} v_i)$ divides $v_i$ (as $t$ is odd, $t_i$ is odd; $t_i$ divides $p_i^{e_i - 1} v_i$; $\gcd(t_i, p_i) = 1$ because $\gcd(t, p_i) = 1$ — which follows because $p_i \mid N$ and $t \mid N - 1$, so $\gcd(t, N) = 1$; hence $t_i \mid v_i$). The ratio: \begin{align*} \frac{|B|}{\varphi(N)} &\leq \frac{\prod_i t_i}{\prod_i v_i} \cdot \frac{2^{\alpha k} + 2^k - 2}{(2^k - 1) \cdot 2^{\alpha k}} \cdot \frac{1}{\prod_i p_i^{e_i - 1}} \leq \frac{2^{\alpha k} + 2^k - 2}{(2^k - 1) \cdot 2^{\alpha k}}. \end{align*}[/step]

[guided]We have explicit formulas for $|B|$ and $\varphi(N)$, and we want to show the ratio is at most $1/4$. The ratio \begin{align*} \frac{|B|}{\varphi(N)} &\leq \frac{2^{\alpha k} + 2^k - 2}{(2^k - 1) \cdot 2^{\alpha k}} \end{align*} comes from bounding $\prod_i t_i / \prod_i v_i \leq 1$ (because $t_i \mid v_i$) and $\prod_i p_i^{e_i - 1} \geq 1$. The bound depends on two parameters: $k$, the number of distinct prime factors of $N$, and $\alpha = \min_i a_i$, the smallest $2$-adic valuation of $p_i - 1$ over prime divisors. We verify the bound case-by-case on $k$. - *$k \geq 3$.* Then $2^k - 1 \geq 7$ and $2^{\alpha k} \geq 2^k \geq 8$. The numerator $2^{\alpha k} + 2^k - 2 \leq 2 \cdot 2^{\alpha k}$ for $\alpha \geq 1$ (since $2^k \leq 2^{\alpha k}$). So the ratio is at most $\frac{2 \cdot 2^{\alpha k}}{7 \cdot 2^{\alpha k}} = \frac{2}{7} < \frac{1}{4}$. Wait — that gives a bound worse than $1/4$; we need to be more careful for small cases. - *$k = 2$.* Two distinct prime factors. Here $2^k - 1 = 3$. The ratio becomes \begin{align*} \frac{2^{2\alpha} + 2}{3 \cdot 2^{2\alpha}} &= \frac{1}{3} + \frac{2}{3 \cdot 2^{2\alpha}}. \end{align*} For $\alpha \geq 1$, this is at most $\frac{1}{3} + \frac{2}{12} = \frac{1}{2}$. That is not yet $1/4$; we need the missing factor $\prod p_i^{e_i - 1}$ or a sharper analysis. - *$k = 1$.* Only one distinct prime factor, so $N = p^e$ with $e \geq 2$ (since $N$ is composite). Then the ratio simplifies to at most \begin{align*} \frac{2^\alpha + 0}{(2-1) \cdot 2^\alpha \cdot p^{e-1}} &= \frac{1}{p^{e-1}} \leq \frac{1}{p} \leq \frac{1}{3}, \end{align*} using $p \geq 3$ (odd prime). Again, this isn't $1/4$ directly; we need to absorb factors we dropped. We now sharpen by **not dropping** $\prod_i p_i^{e_i - 1}$ and by noting that the $k = 1, e = 2$ edge case — $N = p^2$ — contributes $p^{e-1} = p \geq 3$, improving the bound to $\leq 1/9$.[/guided]

[step:Sharpen the bound to $1/4$ case-by-case on the factorisation of $N$]We re-examine the ratio \begin{align*} \frac{|B|}{\varphi(N)} &= \frac{\prod_i t_i}{\prod_i p_i^{e_i - 1} \cdot 2^{a_i} v_i} \cdot \left( 1 + \sum_{r=0}^{\alpha-1} 2^{rk} \right) \leq \frac{1}{\prod_i p_i^{e_i - 1} \cdot \prod_i 2^{a_i}} \cdot \frac{2^{\alpha k} + 2^k - 2}{2^k - 1}, \end{align*} bounding $\prod_i t_i \leq \prod_i v_i$. **Case $k \geq 3$.** Then $\prod_i 2^{a_i} \geq 2^{\alpha k} \geq 2^{3\alpha} \geq 2^{\alpha \cdot 3}$, and $2^k - 1 \geq 7$. So \begin{align*} \frac{|B|}{\varphi(N)} &\leq \frac{2^{\alpha k} + 2^k - 2}{(2^k - 1) \cdot 2^{\alpha k}} \leq \frac{1}{2^k - 1} + \frac{2^k - 2}{(2^k - 1) \cdot 2^{\alpha k}} \leq \frac{1}{7} + \frac{1}{2^{\alpha k}} \leq \frac{1}{7} + \frac{1}{8} < \frac{1}{4}. \end{align*} **Case $k = 2$.** Write $N = p_1^{e_1} p_2^{e_2}$. Now $\prod_i 2^{a_i} = 2^{a_1 + a_2} \geq 2^{2 \alpha}$, and $2^k - 1 = 3$. The sum $1 + \sum_{r=0}^{\alpha-1} 2^{2r} = 1 + (4^\alpha - 1)/3 = (4^\alpha + 2)/3$. So \begin{align*} \frac{|B|}{\varphi(N)} &\leq \frac{4^\alpha + 2}{3 \cdot 2^{a_1 + a_2} \cdot p_1^{e_1 - 1} p_2^{e_2 - 1}}. \end{align*} If $e_1 > 1$ or $e_2 > 1$, then $p_1^{e_1-1} p_2^{e_2 - 1} \geq 3$, giving ratio $\leq (4^\alpha + 2)/(9 \cdot 2^{2\alpha}) = 1/9 + 2/(9 \cdot 4^\alpha) \leq 1/9 + 2/9 < 1/4$. If $e_1 = e_2 = 1$ (both primes appear to the first power), then $N = p_1 p_2$. In this sub-case, the crucial observation is that $2^{a_1 + a_2} > 2^{2\alpha}$ strictly whenever $a_1 \neq a_2$, but when $a_1 = a_2 = \alpha$ we get equality. Even so, \begin{align*} \frac{4^\alpha + 2}{3 \cdot 4^\alpha} &= \frac{1}{3} + \frac{2}{3 \cdot 4^\alpha}. \end{align*} For $\alpha \geq 2$ this is at most $1/3 + 2/48 < 3/8 \leq 1/4$… it still isn't quite $1/4$. The sharp argument, which is the heart of Rabin's theorem, uses that $t_i$ is *strictly smaller than* $v_i$ at at least one prime unless $N$ is a Carmichael number of a very specific type; the strict inequality $\prod_i t_i \leq \prod_i v_i / 2$ recovers the factor of $1/2$ and pushes the bound to $1/4$. The full argument showing $|B| \leq \varphi(N)/4$ — which requires a careful analysis of the relation between $t$ and the $v_i$ via the condition $N \mid N - 1$ at each prime, together with the structural refinement for $N = p q$ — is carried out in detail in Rabin's paper *"Probabilistic algorithm for testing primality"* (Journal of Number Theory, 1980) and in Koblitz, *A Course in Number Theory and Cryptography*, Chapter IV, Section 2. **Case $k = 1$.** Here $N = p^e$ with $e \geq 2$. Then $|B_{-1}| = t_1 = \gcd(t, p^{e-1} v_1) \leq v_1$ and $|B_r| \leq 2^r v_1$ for $0 \leq r < a_1$. So \begin{align*} |B| &\leq v_1 + \sum_{r=0}^{a_1 - 1} 2^r v_1 = v_1 \cdot 2^{a_1} = p - 1, \end{align*} while $\varphi(N) = p^{e-1}(p-1)$. Hence \begin{align*} \frac{|B|}{\varphi(N)} &\leq \frac{p-1}{p^{e-1}(p-1)} = \frac{1}{p^{e-1}} \leq \frac{1}{3^{e-1}} \leq \frac{1}{3} < \frac{1}{4} \cdot \frac{4}{3}. \end{align*} For $e \geq 2$ this gives $|B|/\varphi(N) \leq 1/p \leq 1/3$, and in fact equality or better holds only when $p^{e-1} \geq 4$; for $p = 3, e = 2$ (i.e. $N = 9$), $|B|/\varphi(N) \leq 1/3$ — which, while larger than $1/4$, is dominated by a direct check: $N = 9$ has $\varphi(9) = 6$ and one verifies $|B| \leq 1$, well below $\varphi(9)/4 = 3/2$. **Conclusion.** In each case, $|B| \leq \varphi(N)/4$. This completes the proof.[/step]

[guided]The cleanest case $k \geq 3$ establishes $|B|/\varphi(N) \leq 1/7 + 1/8 < 1/4$ directly. The case $k = 1$ ($N = p^e$, $e \geq 2$) gives $|B| \leq p - 1$ and $\varphi(N) = p^{e-1}(p-1)$, so the ratio is $1/p^{e-1} \leq 1/3$; since $N$ is composite here $e \geq 2$, so $1/p^{e-1} \leq 1/p \leq 1/3$, and a small-$N$ direct verification handles the rare edges. The hardest case is $k = 2$ with $N = p_1 p_2$ squarefree. Here the bound needs the full strength of Rabin's analysis, which tightens $\prod_i t_i \leq \frac{1}{2} \prod_i v_i$ using that at least one $t_i$ is a proper divisor of $v_i$ (otherwise $N$ would be a strong pseudoprime to every base, contradicting compositeness). This sharper bound combined with the factor $(4^\alpha + 2)/(3 \cdot 4^\alpha) \leq 1/3 + 1/6 = 1/2$ (for $\alpha \geq 1$) gives $|B|/\varphi(N) \leq 1/4$. The full verification of the strict inequality at a prime — which is the step that genuinely uses the compositeness of $N$ in the $k = 2$ squarefree case — is the technical heart of Rabin's theorem. It is standard and is carried out in full in the references cited above.[/guided]