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Square expressions (Posted on 2017-02-18) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
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solution Comment 3 of 3 |
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). 


  Posted by xdog on 2017-02-20 20:45:13
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