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Erdős problem 768

Let A(x)A(x) count n≤xn \le x such that every prime p∣np \mid n has a divisor d>1d > 1 of nn with d≡1(modp)d \equiv 1 \pmod p. Erdos asked whether A(x)/x=exp⁡(−(c+o(1))log⁡xlog⁡log⁡x)A(x)/x = \exp(-(c+o(1))\sqrt{\log x}\log\log x). It does, with c=1/(2log⁡2)c = 1/(2\sqrt{\log 2}).

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