Question 567318: Find, with proof, all the perfect squares each of which is the product of four consecutive odd natural numbers. (if there is no such combination then show that through a simple proof)
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Let k-3, k-1, k+1, k+3 be the four numbers. Then for some integer m,
(k-1)(k+1)(k+3))
(k^2 - 1))
Compare with ^2 = k^4 - 10k^2 + 25) . This is definitely a perfect square, so we need this to differ with m^2 by 16. A simple check shows that {9,25} and {0,16} are the only pairs of perfect squares that differ by 16. {0,16} is impossible because we want our perfect square to be odd. If we have {9,25} then
^2 = k^4 - 10k^2 + 25 = 25)
Hence k^4 - 10k^2 = 0, (k^2)(k^2 - 10) = 0. The only integer solution is k = 0, which yields {-3,-1,1,3}, and m^2 = 9. However, this is not in the set of natural numbers, so there are no solutions.
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