What are the positive integers
k and m,
if both k^2 + 4m and m^2 + 5k are squares of integers?
P&p solution to list all the possible couples.
Write k^2 + 4m = (k+x)^2. Then 4m = 2kx + x^2 and x = even, say x = 2a, giving m = ak + a^2.
Similarly m^2 + 5k = (m+y)^2 and 5k = 2my + y^2.
Multiply the first result by 5
5m = 5ak + 5a^2
and the second result by a
5ak = 2amy + ay^2
Substitute for 5ak and solve for m.
m(5 - 2ay) = ay^2 + 5a^2
(5 - 2ay) will be positive, so (a,y) = (1,1), (1,2), or (2,1) corresponding to solutions (k,m) = (1,2), (8,9), (9,22).
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Posted by xdog
on 2017-02-20 20:45:13 |
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