# [seqfan] Sequence of solutions of : x^n + y^(n+1) = z^(n+2)

Jacques Tramu jacques.tramu at echolalie.com
Mon Mar 26 23:23:23 CEST 2018

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Consider the equation
x^n + y^(n+1) = z^(n+2) , x,y,z,n integers >= 1 (1)
and the sequence
s(n) = minimal x solution of (1)

Looking for 'small' solutions x = 2^p*3^q,  one finds :
- n odd : x =   2 ^ ((n+1)(n+1)/2) ,           y = 2 ^ (n(n+1)/2)
- n even : x =  2 ^ (n*(n+1)/2) * 3^(n+2), y = 2^((n^2)/2) * 3^(n+1), z
= 2^((n^2 - n + 2)/2) * 3^n

giving the sequence :
s* =
4,648,256,746496,262144,13759414272,4294967296,4057816381784064,1125899906842624,19147179916555230117888
,... (2)

Relativity heavy computation gives :
s =
2,27,256,472392,262144,13759414272,4294967296,4057816381784064,1125899906842624,19147179916555230117888
, ...? (3)

Ex :
n = 1        2^1 + 5^2 = 3^3              s(1) = 2
n = 2       27^2 + 18^3 = 9^4           s(2) = 27
n = 3       256^3 + 64^4 = 32^5       s(3) = 256
n = 4      472392^4 + 52488^5 = 8748^6        s(4) = 472392
...

Questions :
- Is s(n) ,as shown in (3), correct ?
- Is there an n0 > 4 such as s(n0) < s*(n0) ?
- Is there an n1 > 2 such as s(n1) is odd ?

Any answer, comment will be welcome.

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