SOLUTION: 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 pro

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,

Compare with

. 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

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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