A284666 List of 3-term arithmetic progressions of coprime positive integers whose product is a square.

1, 1, 1, 1, 25, 49, 18, 25, 32, 1, 841, 1681, 49, 169, 289, 50, 169, 288, 49, 289, 529, 128, 289, 450, 98, 625, 1152, 289, 625, 961, 800, 841, 882, 162, 1681, 3200, 288, 1369, 2450, 529, 1369, 2209, 1, 28561, 57121, 49, 5329, 10609, 961, 1681, 2401, 289, 2809, 5329

Offset: 1

Author

Jonathan Sondow, Mar 31 2017

Keywords

• nonn
• tabf

Comments

This is a 3-column table read by rows a, a+d, a+2*d. Each row has product a square. The rows are ordered by the products. The square roots of the products form A284876, which contains A046176. The pairs a,d form A284874.

Goldbach proved that a product of 3 consecutive positive integers is never a square.

Euler proved that a product of 4 consecutive positive integers is never a square.

Erdos and Selfridge (1975) proved that a product of 2 or more consecutive positive integers is never a square or a higher power.

Saradha (1998) proved that 18, 25, 32 is the only arithmetic progression a, a+d, ..., a+(k-1)*d whose product is a square if a>=1, 1=3 with gcd(a,d)=1. In 1997 she showed that the product is not a square or a higher power if a>=1, 1=3 with gcd(a,d)=1.

(1, 1+d, 1+2*d) is in the table if and only if d is in A078522. - Robert Israel, Apr 05 2017 - Jonathan Sondow, Apr 06 2017

Examples

			18*(18+7)*(18+2*7) = 18*25*32 = 9*25*64 = (3*5*8)^2 and gcd(18,25,32) = 1, so 18,25,32 is in the sequence.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..1248 (triples with product < 10^18)
• P. Erdős and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math., 19 (1975), 292-301.
• N. Saradha, On perfect powers in products with terms from arithmetic progressions, Acta Arith., 82 (1997), 147-172.
• N. Saradha, Squares in products with terms in an arithmetic progression, Acta Arith., 86 (1998), 27-43.

Crossrefs

Cf. A046176, A078522, A284874, A284876.

Programs

Maple

N:= 10^11: # to get all triples where the product <= N
Res:= [1,0]:
for a from 1 to floor(N^(1/3)) do
  for d from 1 while a*(a+d)*(a+2*d) <= N do
     if igcd(a,d) = 1 and issqr(a*(a+d)*(a+2*d)) then
       Res:= Res, [a,d]
     fi
  od
od:
Res:= sort([Res], (s,t) -> s[1]*(s[1]+s[2])*(s[1]+2*s[2]) <= t[1]*(t[1]+t[2])*(t[1]+2*t[2])):
map(t -> (t[1],t[1]+t[2],t[1]+2*t[2]), Res); # Robert Israel, Apr 05 2017

Mathematica

nn = 50000; t = {};
p[a_, b_, c_] := a *b*c; Do[
If[p[a, a + d, a + 2 d] <= 2 nn^2 && GCD[a, d] == 1 &&
   IntegerQ[Sqrt[p[a, a + d, a + 2 d]]],
  AppendTo[t, {a, a + d, a + 2 d}]], {a, 1, nn}, {d, 0, nn}];
Sort[t, p[#1[[1]], #1[[2]], #1[[3]]] <
    p[#2[[1]], #2[[2]], #2[[3]]] &] // Flatten

Formula

a(3*k+1) = A284874(2*k+1) and a(3*k+2) = A284874(2*k+1)+A284874(2*k+2) and a(3*k+3) = A284874(2*k+1)+2*A284874(2*k+2) and a(3*k+1)*a(3*k+2)*a(3*k+3) = A284876(k+1)^2 for k>=0.