Solve A^5 – B^5 = C^3 – D^3 = E^2 – F^2 = N for positive integers

 

                   A^5 \; - \; B^5 \; = \; C^3 \; - D^3 \; = \; E^2 \; - \; F^2 \; = \; N
 

A,   B,   C,   D,   E,   F,   and   N   are positive integers

 
for example,
 

Diff 5th,3rd,2nd A1

 

Find other solutions for   A < 100
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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4 Responses to Solve A^5 – B^5 = C^3 – D^3 = E^2 – F^2 = N for positive integers

  1. pipo says:

    Found some more:
    8^5 – 1^5 = 32^3 – 1^3 = 184^2 – 33^2 = 32767
    8^5 – 1^5 = 32^3 – 1^3 = 544^2 – 313^2 = 32767
    8^5 – 1^5 = 32^3 – 1^3 = 2344^2 – 2337^2 = 32767
    14^5 – 7^5 = 161^3 – 154^3 = 931^2 – 588^2 = 521017
    14^5 – 7^5 = 161^3 – 154^3 = 1309^2 – 1092^2 = 521017
    24^5 – 3^3 = 220^3 – 139^3 = 2841^2 – 330^ 2 = 7962381
    27^5 – 8^5 = 243^3 – 32^3 = 3790 ^2 – 219^2 = 14316139

    pipo

  2. pipo says:

    second line should be:
    8^5 – 1^5 = 32^3 – 1^3 = 544^2 – 513^2 = 32767

  3. pipo says:

    And some more:
    64^5 – 27^5 = 1024^3 – 243^3 = 32779^2 – 3882^2 = 1059392917
    64^5 – 27^5 = 1024^3 – 243^3 = 33749^2 – 8922^2 =1059392917
    64^5 – 27^5 = 1024^3 – 243^3 = 34939^2 – 12702^2 = 1059392917

    A pattern emerges:
    ( n^3) ^5 – ((n-1)^3)^5 = ( n^5) ^3 – ((n-1)^5)^3 for all positive n.
    And then we have to find a pair of squares.

    pipo

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