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 p^7 + q^3 = r^2 (Posted on 2011-01-09)
Determine all possible triplets (p, q, r) of positive integers that satisfy this equation:

p7 + q3 = r2, whenever gcd (p, q, r) = 1

 No Solution Yet Submitted by K Sengupta No Rating

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 re: sub-spoiler--another, via computer | Comment 2 of 3 |

10   for R=1 to 1000000:T=R*R
20   for P=1 to int(T^(1/7))
30     for Q=1 to int((T-P^7)^(1/3)+0.5)
40       if P^7+Q^3=T then
50        :G=gcd(P,Q):G=gcd(G,R)
60        :print P,Q,R,G
70     next
80   next
90   next

finds

` p       q        r     gcd 1       2       3       1 3       9       54      3 2       17      71      1 8       64      1536    8 8       128     2048    8 9       243     4374    9 10      225     4625    5 9       486     10935   9 15      225     13500   15`

...

so (2,17,71) also fits

 Posted by Charlie on 2011-01-09 18:04:51

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