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Source. The unnumbered observation near the top of printed page 200 (PDF page 4) of Erdős (1974). The paper calls this easy to see without supplying the construction. The following is a complete reconstruction relative to the named classical theorems.

Use the definition of h(n)h(n) in remark_p199.

Statement. For every real BB there is an odd integer n≥3n\ge3 such that h(n)>Bh(n)>B. In fact, for each fixed integer M≥2M\ge2, there are infinitely many such odd nn with h(n)>Mh(n)>M.

Complete proof. Fix M≥2M\ge2, and put

L=8∏ℓ≤Mℓ an odd primeℓ.L=8\prod_{\substack{\ell\le M\\\ell\text{ an odd prime}}}\ell.

Dirichlet's theorem supplies infinitely many primes q≡−1(modL)q\equiv-1\pmod L. Choose any of these with q>Mq>M and q≥7q\ge7. Then q≡7(mod8)q\equiv7\pmod8, so the supplementary law for the Legendre symbol gives (2/q)=1(2/q)=1. For each odd prime ℓ≤M\ell\le M, quadratic reciprocity and q≡−1(modℓ)q\equiv-1\pmod\ell give

(ℓq)=(−1)(ℓ−1)/2(qℓ)=(−1)(ℓ−1)/2(−1ℓ)=1,\left(\frac{\ell}{q}\right) =(-1)^{(\ell-1)/2}\left(\frac q\ell\right) =(-1)^{(\ell-1)/2}\left(\frac{-1}\ell\right)=1,

because (q−1)/2(q-1)/2 is odd. Multiplicativity now implies (a/q)=1(a/q)=1 for every integer 1≤a≤M1\le a\le M: every prime factor of such an aa was included above and q∤aq\nmid a.

Set n=(q−1)/2n=(q-1)/2. This is an odd integer at least 33. Euler's criterion gives an≡1(modq)a^n\equiv1\pmod q for every 2≤a≤M2\le a\le M. Their collective gcd therefore has the prime divisor qq, so h(n)>Mh(n)>M. The infinitely many choices of qq give infinitely many distinct nn, proving the assertion.

A separate numerical correction. The same source paragraph prints h(15)=5h(15)=5 as an example with P(15)=2P(15)=2. The example is incorrect:

215−1=32767,315−1=14348906,2^{15}-1=32767,\qquad 3^{15}-1=14348906,

and the exact identity

4989077⋅32767−11393⋅14348906=14989077\cdot32767-11393\cdot14348906=1

proves h(15)=3h(15)=3. The source's intended qualitative point, that h(n)h(n) can greatly exceed P(n)P(n), nevertheless follows from the theorem above: for every odd nn, P(n)=2P(n)=2, since p−1p-1 is even for every odd prime pp.

Dependencies. Dirichlet's theorem for a reduced residue class modulo a fixed positive integer; quadratic reciprocity and its supplementary laws; multiplicativity of the Legendre symbol; and Euler's criterion. Their proofs are external. remark_p199 establishes that the threshold is finite.

Bears on. #770. Unboundedness along odd exponents does not assert that h(n)h(n) tends to infinity or resolve the question of infinitely many values equal to three.