a^5 – b^5 = c^3 – d^3 = e^2 – f^2

Find solutions to

$a^5 \; - \; b^5 \; = \; c^3 \; - \; d^3 \; = \; e^2 \; - \; f^2$

where   $a, b, c, d, e, f$   are positive integers.

for example,

$4^5 \; - \; 2^5 \; = \; 10^3 \; - \; 2^3 \; = \; 39^2 \; - \; 23^2 \; = \; 992$

Note that there are other solutions for the difference of squares:

$992 \; = \; 66^2 \; - \; 58^2$
$992 \; = \; 126^2 \; - \; 122^2$
$992 \; = \; 249^2 \; - \; 247^2$

Find other solutions

Paul found:

$6^5 \; - \; 3^5 \; = \; 21^3 \; - \; 12^3 \; = \; 87^2 \; - \; 6^2$
$8^5 \; - \; 1^5 \; = \; 32^3 \; - \; 1^3 \; = \; 184^2 \; - \; 33^2$
$9^5 \; - \; 6^5 \; = \; 48^3 \; - \; 39^3 \; = \; 227^2 \; - \; 16^2$
$14^5 \; - \; 7^5 \; = \; 161^3 \; - \; 154^3 \; = \; 931^2 \; - \; 588^2$

math grad - Interest: Number theory
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2 Responses to a^5 – b^5 = c^3 – d^3 = e^2 – f^2

1. paul says:

This formatting might turn out nasty, the format can be seen if you follow the first one against the example.

{4,2} ,
10 2 39 23
10 2 66 58
10 2 126 122
10 2 249 247
992

{6,3} ,
21 12 87 6
21 12 137 106
21 12 153 126
21 12 423 414
21 12 1257 1254
21 12 3767 3766
7533

{8,1} ,
32 1 184 33
32 1 544 513
32 1 2344 2337
32 1 16384 16383
32767

{9,6} ,
48 39 227 16
48 39 357 276
48 39 963 936
48 39 2853 2844
48 39 8547 8544
48 39 25637 25636
51273

{14,7} ,
161 154 931 588
161 154 1309 1092
161 154 5341 5292
161 154 8419 8388
161 154 37219 37212
161 154 260509 260508
521017

Paul.