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Algebra

Blazing goIITian

Joined:	27 Dec 2010
Post:	1474

11 Feb 2012 23:42:56 IST

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213

good number theory question.

Mathematics

prove that for any integer n > 1,

4^n + n^4 is composite.

it is a rmo question i think. i will post my solution to it too. but i want to see new solutions!!

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Comments (2)

prahlad kumar sharma

Cool goIITian

Joined: 21 Jul 2010

Posts: 88

12 Feb 2012 10:33:23 IST

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when n is even 4^n + n^4 hamesa composite hi hoga yeh toh saf saf dikhtha hain..

so heres the proof when n is odd

let n= 2p+1 ( since n is odd)

now

4^n + n^4

=> (2ⁿ+n²)² - 2⁽ⁿ⁺¹⁾n²

put n = 2p+1

[2^(2p⁺¹⁾ + (2p+1)²]² - 2^(2p+2)(2p+1)²

=> [2^(2p⁺¹⁾ + (2p+1)²]²- [2⁽^p+¹⁾(2p+1)]² , which comes in the form a²-b² which can be written as (a+b)(a-b)

bas bangaya kaam.

jagdish singh

Blazing goIITian

Joined: 19 Jan 2008

Posts: 805

16 Feb 2012 08:52:53 IST

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$\hspace{-16}$Yes Prahalad is absolutely Right. Fantastic Solution Man.......\\\\ \bullet$\;\; $When $\mathbf{n\in 2\mathbb{K},\;\; \mathbb{K}\in \mathbb{N}}$\\\\ Then $\mathbf{n^4+4^n\in 2{\mathbb{K}}}$\;, Means always even $\mathbf{>2}$.\\\\ So It is a Composite no.\\\\ \bullet$\;\; $When $\mathbf{n\in 2\mathbb{K}+1,\;\; \mathbb{K}\in \mathbb{N}}$\\\\ Then $\mathbf{n^4+4^{2K+1}=n^4+4.(2)^{4k}}$\\\\ Let $\mathbf{(2)^{k}=x}$, Then\\\\ $\mathbf{n^4+4^{2k+1}=n^4+4x^4=\left(n^2+2x^2+2nx\right).\left(n^2+2x^2-2nx\right)}$\\\\ Which is again Composite no.\\\\ Using Sophie German Identitie...\\\\ $\boxed{\boxed{\mathbf{a^4+4b^4=\left(a^2+2b^2+2ab\right).\left(a^2+2b^2-2ab\right)}}}$$

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