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Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).
5

%I #16 Jun 21 2020 11:50:30

%S 5,7,8,10,12,13,14,17,20,25,26,28,29,32,33,34,37,40,41,45,48,50,52,53,

%T 56,57,58,61,63,65,68,71,72,73,74,78,80,82,85,89,90,97,98,100,101,104,

%U 105,106,109,112,113,114,116,117,122,125,126,128

%N Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

%C Motivated by the search of solutions to a^n + b^(2n+2)/4 = (perfect square), which arises when searching solutions to x^n + y^(n+1) = z^(n+2) of the form x = a*z, y = b*z. It turns out that many solutions are of the form a^n = d (b^(n+1) + d), where d is a perfect power.

%H Robert Israel, <a href="/A304433/b304433.txt">Table of n, a(n) for n = 1..5970</a>

%e 5^3 = 125 = 4^2 + 11^2; 7^3 = 10^2 + 3^5; 8^3 = 13^2 + 7^3, ...

%p N:= 200: # to get terms <= N

%p N3:= N^3:

%p P:= {seq(seq(x^k, k=3..floor(log[x](N3))), x=2..N)}:

%p filter:= proc(n) local n3, Pp, x, y;

%p n3:= n^3;

%p if remove(t -> subs(t, x)<=1 or subs(t, y)<=1 or subs(t, x-y)=0, [isolve(x^2+y^2=n3)]) <> [] then return true fi;

%p Pp:= map(t ->n3-t, P minus {n3, n3/2});

%p (Pp intersect P <> {}) or (select(issqr, Pp) <> {})

%p end proc:

%p select(filter, [$2..N]); # _Robert Israel_, Jun 01 2018

%t M = 200;

%t M3 = M^3;

%t P = Union@ Flatten@ Table[Table[x^k, {k, 3, Floor[Log[x, M3]]}], {x, 2, M}];

%t filterQ[n_] := Module[{n3, Pp, x, y}, n3 = n^3; If[Solve[x > 1 && y > 1 && x != y && x^2 + y^2 == n3, {x, y}, Integers] != {}, Return[True]]; Pp = n3 - (P ~Complement~ {n3, n3/2}); (Pp ~Intersection~ P) != {} || Select[ Pp, IntegerQ[Sqrt[#]]&] != {}];

%t Select[Range[2, M], filterQ] (* _Jean-François Alcover_, Jun 21 2020, after _Robert Israel_ *)

%o (PARI) L=200^3; P=List(); for(x=2, sqrtnint(L,3), for(k=3, logint(L, x), listput(P, x^k))); #P=Set(P) \\ This P = A076467 \ {1} = A111231 \ {0} up to limit L.

%o is(n,e=3)={for(i=1, #s=sum2sqr(n=n^e), vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=1, #P, n>P[i]||return; ispower(n-P[i])&& P[i]*2 != n && return(1))} \\ The above P must be computed up to L >= n^3. For sum2sqr() see A133388.

%Y Cf. A001597 (perfect powers), A076467 (cubes and higher powers), A304434, A304435, A304436 (analog for n^4, n^5, n^6).

%K nonn

%O 1,1

%A _M. F. Hasler_, May 12 2018