A307745 -id:A307745 - cp's OEIS frontend

A208242 Perfect powers y^q with y > 1 and q > 1 which are Brazilian repunits with three or more digits in some base.

121, 343, 400

Offset: 1

Bernard Schott, Jan 11 2013

These three numbers are the only known solutions y^q of the Nagell-Ljunggren equation (b^m-1)/(b-1) = y^q with y > 1, q > 1, b > 1, m > 2. Yann Bugeaud and Maurice Mignotte propose two alternative conjectures:
A) The Nagell-Ljunggren equation has only these three solutions.
Considering the current state of our knowledge, this conjecture seems too ambitious, while the next one seems more reasonable.
B) The Nagell-Ljunggren equation has only a finite number of solutions.
This last conjecture is true if the abc conjecture is true (see article Bugeaud-Mignotte in link, p. 148).
Consequence: 121 is the only known square of prime which is Brazilian.
There are no other solutions for some base b < 10000.
Some theorems and results about this equation:
With the exception of the 3 known solutions,
1) for q = 2, there are no other solutions than 11^2 and 20^2,
2) there is no other solution if 3 divides m than 7^3,
3) there is no other solution if 4 divides m than 20^2. - Bernard Schott, Apr 29 2019
From David A. Corneth, Apr 29 2019: (Start)
Intersection of A001597 and A053696.
a(4) > 10^25 if it exists using constraints above.
In the Nagell-Ljunggren equation, we need b > 2. If b = 2, we get y^q = 2^m - 1 which by Catalan's conjecture has no solutions (see A001597). (End)

			121 = 11^2 =  (3^5 - 1)/ (3 - 1) = 11111_3.
343 =  7^3 = (18^3 - 1)/(18 - 1) =   111_18.
400 = 20^2 =  (7^4 - 1)/ (7 - 1) =  1111_7.

Y. Bugeaud and M. Mignotte, L'équation de Nagell-Ljunggren (x^n-1)/(x-1) = y^q, Enseign. Math. 48(2002), 147-168.

Cf. A001597, A053696, A220571 (Brazilian composites), A307745 (similar but with digits > 1).

is(n) = if(!ispower(n), return(0)); for(b=2, n-1, my(d=digits(n, b)); if(#d > 2 && vecmin(d)==1 && vecmax(d)==1, return(1))); 0 \\ Felix Fröhlich, Apr 29 2019

Small edits to the name by Bernard Schott, Apr 30 2019

A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

Original entry on oeis.org

3, 39, 543, 7563, 105339, 1467183, 20435223, 284625939, 3964327923, 55215964983, 769059181839, 10711612580763, 149193516948843, 2077997624703039, 28942773228893703, 403120827579808803, 5614748812888429539, 78203362552858204743, 1089232326927126436863, 15171049214426911911339

Offset: 1

Bernard Schott, Feb 17 2021

nonn

Corresponding x are in A028231.
This equation belongs to the family of equations studied by Kustaa A. Inkeri, y^m = a * (x^q-1)/(x-1) with here: m=2, a=3, q=3. This equation is exhibed in A307745 by Giovanni Resta to prove that this sequence has infinitely many terms.
This Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions because the Pell-Fermat equation u^2 - 3*v^2 = -2 also has infinitely many solutions. The corresponding (u,v) are in (A001834, A001835) and for each pair (u,v), the corresponding solutions of 3*(x^2+x+1) = y^2 are x = (3*u*v-1)/2 and y = 3*(u^2+1)/2.
Note that if y = 3*z, this equation becomes 3*z^2 = x^2+x+1 with solutions (x, z) = (A028231, A001570).

			The first few values for (x,y) are (1,3), (22,39), (313,543), (4366,7563), (60817,105339), ...

Kustaa A. Inkeri, On the Diophantine equation a(x^n-1)/(x-1) = y^m, Acta Arithmetica, Vol. 21, No. 1 (1972), pp. 299-311.

Cf. A001834, A001835, A001570, A028231, A307745.

Subsequence of A158235, for a(n)>3.

f[x_] := Sqrt[3*(x^2 + x + 1)]; f /@ LinearRecurrence[{15, -15, 1}, {1, 22, 313}, 20] (* Amiram Eldar, Feb 17 2021 *)

a(n) = 3*A001570(n). - Hugo Pfoertner, Feb 17 2021

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

More terms from Amiram Eldar, Feb 17 2021

cp's OEIS Frontend

A208242 Perfect powers y^q with y > 1 and q > 1 which are Brazilian repunits with three or more digits in some base.

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