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Algebra

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27 Nov 2008 15:31:06 IST

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For 10th or below....

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(i) If $n$ is a positive integer, such that $2n+1$ is a perfect square, show that $n+1$ is the sum of two successive perfect squares.

(ii) If $3n+1$ is a perfect square, show that $n+1$ is the sum of three perfect square.

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Rahul Duggal

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Joined: 10 May 2008

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27 Nov 2008 19:12:19 IST

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the image is getting distorted you can download the picture to see my solution

®µD®A

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27 Nov 2008 23:49:27 IST

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Due to a problem i could not give the solution to the first one...

Here is the second one.

By Euclid'sdivison lemma we know that a number can be in the form of $3q+r, r=\{0,1,2\},\ q,r \in R$

So a square can be in the form of

$(3q)^2=9q^2\\\\(3q+1)^2=9q^2+1+6q=3(3q^2+2q)+1\\\\(3q+2)^2=9q^2+4+12q=3(3q^2+1+4q)+1$

If 3n+1 is a perfect square then

$n=3q^2+2q\\or\ n=3q^2+1+4q$

(CASE 1)

3q+1 is odd

then n+1 is in form: 3q^2 + 4q + 2 = 2 (q+1)^2 + q^2

(CASE 2)

3q+1 is even

then n+1 is in form : 3q^2 + 2q + 1 =(q+1)^2 + 2 q^2

Lakshya Bhardwaj

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Joined: 20 Jul 2008

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29 Nov 2008 15:02:08 IST

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First one:

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