Not another Perfect Square Problem

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Product of any 4 consecutive integers is always 1 less than a perfect square.

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5 Answers

341

Hari Shankar ·2010-08-07 07:14:10

Is everyone on a holiday or what? That's the only explanation for this one to go unanswered.

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62

Lokesh Verma ·2010-08-07 08:15:49

lol.. i guess ;)

atleast the xi class guys should be trying this one ;)

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49

Subhomoy Bakshi ·2010-08-07 15:51:04

n(n+1)(n+2)(n+3)

=(n²+3n)(n²+3n+2)

=n⁴+6n³+9n²+2n²+6n

=n⁴+6n³+11n²+6n+1-1

=(n²)² + (3n)² + (1)² + 2.(n²).(1) + 2.(3n).(1) + 2.(n²).(3n) - 1

=(n² + 3n +1)² - 1

=k² - 1

thus, n(n+1)(n+2)(n+3) is 1 less than a perfect square!! :D

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62

Lokesh Verma ·2010-08-07 15:58:04

n(n+1)(n+2)(n+3)

=(n²+3n)(n²+3n+2)

you made it worse from here

t(t+2) = t²+2t+1-1 = (t+1)²-1

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Subhomoy Bakshi ·2010-08-07 15:59:41

hehe yea....

didi it in that way...

and just b4 completing just forgot why i had grouped it :P

stupid me! :P

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Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

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