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Numbers n such that n^4 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).
5

%I #24 Nov 29 2023 17:19:00

%S 3,5,6,9,10,12,13,14,15,17,20,24,25,26,28,29,30,34,35,36,37,39,40,41,

%T 42,45,48,50,51,52,53,55,57,58,60,61,63,65,68,70,71,72,73,74,75,78,80,

%U 82,85,87,89,90,91,95,96,97,98,100,101,102,104,105,106,109,110,111,113

%N Numbers n such that n^4 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="/A304434/b304434.txt">Table of n, a(n) for n = 1..1301</a>

%e 3^4 = 2^5 + 7^2; 5^4 = 7^2 + 24^2, ...

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

%p N4:= N^4:

%p P:= {seq(seq(x^k,k=3..floor(log[x](N4))),x=2..floor(N4^(1/3)))}:

%p filter:= proc(n) local n4, Pp;

%p n4:= n^4;

%p if remove(t -> subs(t,x)<=1 or subs(t,y)<=1 or subs(t,x-y)=0, [isolve(x^2+y^2=n4)]) <> [] then return true fi;

%p Pp:= map(t ->n4-t, P minus {n4, n4/2});

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

%p end proc:

%p A:= select(filter, [$2..N]); # _Robert Israel_, May 24 2018

%o (PARI) L=200^4; 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_A304434(n)={for(i=1, #s=sum2sqr(n=n^4), 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^4. For sum2sqr() see A133388.

%Y Cf. A304433, A001597 (perfect powers), A076467 (third or higher powers).

%K nonn

%O 1,1

%A _M. F. Hasler_, May 22 2018