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The fourth power of the common differenc...

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive of it. Prove that the resulting sum is the squares of an integer.

Text Solution

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Let a-3d,a-d,a+d and a+3d be four consecutive terms of an A.P. with common difference 2d.
Hence P `=(2d)^(4)+(a-3d)(a-d)(a+d)(a+3d)`
`=16a^(4)+(a^(2)-9d^(2))(a^(2)-d^(2))`
`=(a^(2)-5d^(2))^(2)`
which is integer.
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