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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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 some thoughts Comment 3 of 3 |

Examining the sum of the remainders (p^7+q^3), mod 8 gives:
1. Both even
2. One even, one (8n+1)
3. One(8m+1), one (8n+3)
4. One(8m+1), one (8n+7)
5. One(8m+3), one (8n+5)
6. One(8m+5), one (8n+7)

as the possibilities for {p,q}, but most of these result in c being some nasty-looking root.

We can ignore 'both even' by virtue of the stipulation of the problem.

This leaves 2. One even, one (8n+1) where:

p=2m,q=8n+1; c^2=8(16m^7+64n^3+24n^2+3n)+1{m,n,c} {0,0,1}{0,1,27}{1,2,71}{0,3,125}{0,6,343} where for higher values m has to be 0.

p=8n+1,q=2m
c^2=8(m^3+262144n^7+229376n^6+86016n^5+ 17920n^4+2240n^3+168n^2+7n)+1
{m,n,c} {0,0,1}{1,0,3}{0,1,2187}{0,3,78125} where for higher values m again has to be 0.

There is also the interesting case:

p=(a^2-1),q=(a-1)^2*(a+1)^2
((a^2-1)^3)^2*(a^2)=c^2
where there is  a solution for every positive integer a, however p then divides q, again contrary to the stipulation of the problem.

So the valid solutions are:
p=1,q=2,r=3
p=2,q=17,r=71

Edited on January 10, 2011, 4:33 am
 Posted by broll on 2011-01-10 04:16:12

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