## 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,

Find other solutions for   A < 100

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