Egyptian fractions with restrictions (Q2893036)

Consider the very well-known Diophantine equation NEWLINE\[NEWLINE\sum_{i=1}^k 1/x_{i}=1,\tag{*}NEWLINE\]NEWLINE in integers \( 1< x_1<\ldots <x_k \). Fix the distinct primes \(p_1,\ldots,p_t \) and denote by \( S(p_1,\ldots,p_t) \) the set of positive integers all prime divisors of which belong to \( \{p_1,\ldots,p_t\} \). This paper deals with those solutions \((x_1,\ldots, x_k) \) of (*) which satisfy \(x_i\in S(p_1,\ldots,p_t)\) for every \(i=1,\ldots,k\); let \(T_k(p_1,\ldots,p_k)\) be the number of these solutions. Obviously, the relation \(T_k(p_1,\ldots,p_k)\geq 1\) means that (*) is solvable in integers \(x_1,\ldots,x_k\) belonging to \( S(p_1,\ldots,p_t) \). Clearly, one cannot expect that this relation is valid for every \(k\), therefore, denote by \(K(p_1,\ldots,p_t)\) the set of all \(k\)'s such that \(T_k(p_1,\ldots,p_k)\geq 1\). The authors prove that \(K(p_1,\ldots,p_t)\) is the union of finitely many arithmetic progressions. Further, they prove that there exist constants \(k_0=k_0(p_1,\ldots,p_t)\) and \(c_1=c_1(p_1,\ldots,p_t)>1\), such that, for all \(k>k_0\) with \(k\in K(p_1,\ldots,p_t)\), it is true that NEWLINE\[NEWLINE c_1^k \leq T_k(p_1,\ldots,p_t) \leq \sqrt{2}\,^{tk^2(1+o_k(1))}. NEWLINE\]NEWLINE Another interesting result of the paper is that \(K(p_1,\ldots,p_t)\) is non-empty, that is, for some \(k>1\), equation (*) is solvable in integers \(x_1,\ldots,x_k \in S(p_1,\ldots,p_t) \), if and only if NEWLINE\[NEWLINE \prod_{i=1}^t{p_i\over p_i-1} > 2. NEWLINE\]NEWLINE For the first few primes the authors prove the following results:NEWLINENEWLINE(a) \( T_k(3,5,7)\geq c_1\sqrt{62}\,^k\) for a computable constant \(c_1\) and any odd integer \(k\geq 11\).NEWLINENEWLINE(b) \( T_k(2,3,5)\geq c_2\sqrt{368}\,^k\) for a computable constant \(c_2\) and any odd integer \(k\geq 3\).NEWLINENEWLINEA relevant result of this paper concerning equation (*) when \( 1< x_1<\ldots <x_k \) (without any other restriction on the \(x_i\)'s) is that, when \(k=2n+1\) with \(n\geq 4\), the number of solutions is at least \(\sqrt{2}\,^{(n+1)(n-4)}\) (easily, if \(k\) is even, then the equation is impossible).