Guest

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.

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.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
Let a – 3d, a – d, a + d and a + 3d be any four consecutive terms of an A. P. with common difference 2d ∵ Terms of A. P. are integers, 2d is also an integer.
Hence p = (2d)4 + (a – 3d) (a – d) (a + d) (a + 3d)
= 16d4 + (a2 – 9d2 ) (a2 – d2) = a2 – 5d2)2
Now, a2 – 5d2 = a2 – 9d2 + 4d2
= (a – 3 d) (a + 3 d) + (2d)2 = some integer
Thus p = square of an integer.

Thanks
Aditi Chauhan
askIITians Faculty

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free