Simple! But is it really?

Algebra Level 2

p 3 + ( e τ i ) ( e i π ) 2 = q 2 \large p^3 + \frac{(e^ {\tau i})-(e^{-i \pi})}{2} = q^2

Given that p p and q q are positive integers, what is the value of q p q^p ?


The answer is 9.

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2 solutions

It must be known that τ = 2 π \tau =2\pi , and thus e τ i = e 2 π i e^{ \tau i} = e^{2\pi i} . As well as Euler's formula: e x i = c o s ( x ) + i s i n ( x ) e^{x i} = cos(x) + i sin(x)
Using Euler's formula to compute e 2 π i e π i e^{2 \pi i}-e^{- \pi i} :
e 2 π i e π i = ( c o s ( 2 π ) + i s i n ( 2 π ) ) ( c o s ( π ) + i s i n ( π ) ) = ( 1 ) ( 1 ) = 2 e^{2 \pi i}-e^{- \pi i} = (cos( 2 \pi ) +i sin( 2 \pi ))-(cos(- \pi ) +i sin( - \pi )) =(1)-(-1) =2
Thus p 3 + e 2 π i e π i 2 = p 3 + 2 2 = p 3 + 1 p^{3} + \frac{e^{2 \pi i}-e^{- \pi i}}{2} = p^{3} + \frac{2}{2} = p^{3} +1 The only solution to p 3 + 1 = q 2 p^{3} +1 = q^{2} where both p p and q q are integers is p = 2 p=2 and q = 3 q=3 ,
It follows then that q p = 3 2 = 9 q^{p} = 3^{2} = \textbf{9}


Mohammad Farhat
Sep 1, 2018

I confess, I complicated the problem for no reason by bringing in e e , τ \tau , π \pi & i i .

We know that ( e τ i ) = ( e 2 π i ) = ( 1 ) (e^ {\tau i}) = (e^ { 2\pi i}) = (1) and ( e i π ) = ( 1 ) (e^{-i \pi}) = (-1) . Following that, ( 1 ) ( 1 ) = 2 (1)-(-1)=2 and 2 2 = 1 \dfrac{2}{2} = 1

So the simplified version would look something like :

p 3 + 1 = q 2 \huge p^3 + 1 = q^2

And by Catalan's Conjecture, the only positive integer solutions are \text {q = 3 & p = 2 }

Please say positive integers instead of integers, because ( p , q ) = ( 0 , 1 ) , ( 0 , 1 ) , ( 1 , 0 ) (p,q)=(0,1), (0,-1), (-1,0) are also integer solutions.

X X - 2 years, 9 months ago

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I added whole numbers and Thank You!

Mohammad Farhat - 2 years, 9 months ago

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Whole numbers includes 0, so (0,1) is still possible.

X X - 2 years, 9 months ago

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@X X Ok! Sorry!

Mohammad Farhat - 2 years, 9 months ago

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