Question

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Answer 1

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 resulting sum is the square of an integer.

Answer 2

@ganeshie8

Answer 3

i think i got it wait

Answer 4

so quick ?

Answer 5

Is the square of something
\[a ^{4}+6a ^{3}d+6a ^{2}d ^{2}+5a ^{2}d+6ad ^{3}+d ^{4}\]

Answer 6

this *

Answer 7

@ganeshie8

Answer 8

how did u prove ?

Answer 9

But is it the square of something

Answer 10

r u thre

Answer 11

yes so are you trying to prove below is a perfect square ?

\[d^4 + a(a+d)(a+2d)(a+3d)\]

Answer 12

yup

Answer 13

it is indeed a perfect square
http://www.wolframalpha.com/input/?i=factor+d%5E4+%2B+a%28a%2Bd%29%28a%2B2d%29%28a%2B3d%29

Answer 14

How to know it without wolfram

Answer 15

atleast it doesn't look straightforward to me..

Answer 16

Is there no method

Answer 17

this looks simpler, check :
http://www.askiitians.com/forums/Algebra/the-fourth-power-of-the-common-difference-of-an-ar_104089.htm

Answer 18

nice logic thanks