# 2 mark proof question (1 Viewer)

#### oredbayz

##### New Member
RTP: 2^n + 3^n =/= 5^n

is it possible to do by induction? any help appreciated

thanks

#### oredbayz

##### New Member
just had another thought, would this work?

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#### notme123

##### Well-Known Member
it was 2 marks and it didnt ask induction. i think theres another intended method. would anyone happen to have a solution that is equaivalent to 2 marks

#### Trebla

$\bg_white 5^n = (2+3)^n \\\\= \sum_{k=0}^n\binom{n}{k}2^{n-k}3^{k}\\\\=2^n+\sum_{k=1}^{n-1}\binom{n}{k}2^{n-k}3^{k}+3^n$

The middle term is non-zero/positive provided n>1 as an integer so hence they cannot be equal.

#### CM_Tutor

##### Moderator
Moderator
Can't you just prove this by contradiction?

\bg_white \begin{align*} \text{Suppose that the theorem is false, and so that}\ 2^n + 3^n &= 5^n \\ 5^n - 2^n - 3^n &= 0 \\ \text{However, LHS}\ &= (2 + 3)^n - 2^n - 3^n \\ &= \binom{n}{0} 2^n + \binom{n}{1}2^{n-1} \times 3 + \binom{n}{2}2^{n-1} \times 3^2 + ... + \binom{n}{n-1}2^1 \times 3^{n - 1} + \binom{n}{n} 3^n - 2^n - 3^n \qquad \text{provided n \in \mathbb{Z} and n > 1} \\ &= 2^n + \binom{n}{1}2^{n-1} \times 3 + \binom{n}{2}2^{n-1} \times 3^2 + ... + \binom{n}{n-1}2^1 \times 3^{n - 1} + 3^n - 2^n - 3^n \qquad \text{as \binom{n}{0} = \binom{n}{n} = 1} \\ &= \binom{n}{1}2^{n-1} \times 3 + \binom{n}{2}2^{n-1} \times 3^2 + ... + \binom{n}{n-1}2^1 \times 3^{n - 1} \\ &> 0 \qquad \text{as every term in this sum is positive} \\ &\neq \text{RHS} \end{align*}

So, we have a contradiction provided $\bg_white n \geqslant 2$, and so the assumption / supposition is false and the theorem is true.

Hence, $\bg_white 2^n + 3^n \neq 5^n$ for any integer $\bg_white n \geqslant 2$. The statement $\bg_white 2^n + 3^n = 5^n$ is obviously true for $\bg_white n = 1$

#### CM_Tutor

##### Moderator
Moderator
I wonder what they would make of a proof that invoked Fermat's Last Theorem and noted that this was proven by Andrew Wiles a few years ago, pointing at that if there are no solutions in positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$ for the statement $\bg_white a^n + b^n = c^n$ for any integer $\bg_white n > 2$, and then pointing out that with $\bg_white a = 2$, $\bg_white b = 3$, and $\bg_white c = 5$, the $\bg_white n = 2$ case is also false as $\bg_white 2^2 + 3^2 = 4 + 9 = 13 \neq 25 = 5^2$. Thus $\bg_white 2^n + 3^n \neq 5^n$ for all integers $\bg_white n > 1$.

#### CM_Tutor

##### Moderator
Moderator
Actually, proving the more general result that, for positive integers $\bg_white a$ and $\bg_white b$, and for all integers $\bg_white n > 1$

$\bg_white a^n + b^n \neq (a + b)^n$

might have been easier.

#### 5uckerberg

##### Well-Known Member
Will there be anybody here showing Sydney Morning Herald how to do Q16 this year?

I am curious. There will always be those boomers who say so what, skills in the real world are more important than a little mathematical equation. Typical boomer talk from the Sydney Morning herald. They never learn.

##### -insert title here-
For a non-binomial method, divide both sides by 3^n to obtain

(2/3)^n + 1 = (5/3)^n

The left hand side has a base smaller than 1 and is a strictly decreasing function of n, and the right hand side is likewise strictly increasing.

Thus, there can only be at most one solution, which is obviously n=1. Hence, the two expressions are unequal for n>1

##### -insert title here-
I wonder what they would make of a proof that invoked Fermat's Last Theorem and noted that this was proven by Andrew Wiles a few years ago, pointing at that if there are no solutions in positive integers $\bg_white a$, $\bg_white b$, and $\bg_white c$ for the statement $\bg_white a^n + b^n = c^n$ for any integer $\bg_white n > 2$, and then pointing out that with $\bg_white a = 2$, $\bg_white b = 3$, and $\bg_white c = 5$, the $\bg_white n = 2$ case is also false as $\bg_white 2^2 + 3^2 = 4 + 9 = 13 \neq 25 = 5^2$. Thus $\bg_white 2^n + 3^n \neq 5^n$ for all integers $\bg_white n > 1$.
1995 is more than a few years....

#### jxdesml

##### Member
the mods on here are so smart...

Moderator

#### npine2021

##### New Member
can anyone find the ext 2 paper?

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#### michealhunter@school

##### New Member
I let the LHS and RHS be different functions then I differeniated both functions and showed that 5^x increases faster than 2^x
+ 3^x after x =1. Dunno if its worth full marks but it worked out for me.

#### icycledough

##### Well-Known Member
I let the LHS and RHS be different functions then I differeniated both functions and showed that 5^x increases faster than 2^x
+ 3^x after x =1. Dunno if its worth full marks but it worked out for me.
That does seem like a valid way of going about it (I think there must be various ways to go about it, so as long as you could set your answer out and it made sense in a chronological manner, then it should be fine). I guess with the topics of proofs, what makes it nice is the level of flexibility with regards to what method you can use for an individual question.

#### epicuber7

##### New Member
I factorised 2^n + 3^n as (2+3)(2^(n-1) - 3*2^(n-2) + ... + 3^(n-1)) and assumed the right hand bracket could not be 5^(n-1) rip

#### dwhinc

##### New Member
For n>1,
(2/5)^n + (3/5)^n < 2/5 + 3/5 =1.

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