This observation seems extremely intuitive, but is a lot helpful when dealing with most of the problems involving Second Principle of Mathematical Induction :-

$\boxed{a^{k}-b^{k} = (a^{k-n}-b^{k-n})(a^{n}+b^{n})-(ab)^{n}(a^{k-2n}-b^{k-2n})}$

I suggest the reader to accept the following problem as a challenge and post its solution in the comments ( It was this problem that made me realize about the identity) :-

INMO Type Problem :- Let $R = (5 \sqrt{5} + 11)^{2n+1}$ and let $f$ be the fractional part of $R$. Show that $Rf = 4^{2n+1}$.

By the way the above problem is not original.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest$R=I+f$

$0<5\sqrt 5 -11 <1$

Let $(5\sqrt 5 -11)^{2n+1}$=$f'$

Now , $I+f-f' = (5\sqrt 5+11)^{2n+1}-(5\sqrt 5 -11) ^{2n+1}= 2(^nC_1 (5\sqrt 5)^{2n}11 + ^n C3 (5\sqrt 5)^{2n-2}11^3 .............)$

Here RHS is integer So Lhs should be too so $f=f'$

Hence $Rf=Rf'=((5\sqrt 5 +11)^{2n+1}(5\sqrt 5 -11)^{2n+1}) = 4^{2n+1}$

Log in to reply

How do you say that RHS is an integer ? There is no proper reason for your claim. And also please do not include facts like Binomial Expansion, as the aim of this note is to deal with elegant proofs of theorems that are direct consequences of Induction Axioms.

Log in to reply

RHS is an integer cause $5\sqrt 5$ always have even power and it is old IITJEE problem too.This method is absolutely correct .I subtracted those two terms to make RHS an integer.It is not a claim it is consequence of my subtraction

Log in to reply

Log in to reply

Can u tell me this question came in which yr in INMO.

Log in to reply

Log in to reply

typeproblem, I guess it never actually came. However, I have a book (Problem-Solving StrategiesbyArthur Engel), that has a problem which uses a very similar approach (adding the conjugate of the irrational number) for a certain problem.In 1980, the oncoming IMO in Mongolia was cancelled on political grounds, Luxembourg hosted an Ersatz-IMO in Mersch. According to my book, that problem occured in this

Ersatz-IMO in 1980. This was the problem:Find the first digit before and after the decimal in:${ (\sqrt { 2 } +\sqrt { 3 } ) }^{ 1980 }$Log in to reply

I think there is a fallacy :- $I + f - f' =$

an integerneed not imply that $f = f'$. It may even be like $f = 1 + f'$ or $f = 2 + f'$ or generally $f = n + f'$ where $n$ is any integer, and still the fact that LHS ( as calculated by binomial expansion ) is an integer holds good.Log in to reply

Hey man $f$ is fractional part of x hence $0 \le f<1$.Its a standard solution to these type of problems.

Log in to reply

Log in to reply

Great observation.

The $n = 1$ case is similar to what we get from Newton's Identities, namely

$a^k + b^k = ( a^{k-1} + b^{k-1} ) ( a + b ) - ( ab) ^ 2 ( a ^ {k-2} + b^ { k - 2 } ).$

Log in to reply

Thanks Sir ! There is a really close connection between my observation and Newton's Identities, although they are used for completely different purposes !

Log in to reply