The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.
Text Solution
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(a – 3d) (a – d) (a + d) (a + 3d) + 16d 4 = (a 2 – 9d 2 ) (a 2 – d 2 ) + 16d 4
= a 4 – a 2 d 2 – 9a 2 d 2 + 9d 4 + 16d 4 = a 4 – 10a 2 d 2 + 25d 4 = [a 2 – 5 d 2 ] 2
= (a 2 – d 2 – 4d 2 ) 2 = ( (a – d) (a + d) – (2d) 2 ) 2
(a – d), (a + d), 2d are integers. Hence Proved
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