p 3 + 2 ( e τ i ) − ( e − i π ) = q 2
Given that p and q are positive integers, what is the value of q p ?
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I confess, I complicated the problem for no reason by bringing in e , τ , π & i .
We know that ( e τ i ) = ( e 2 π i ) = ( 1 ) and ( e − i π ) = ( − 1 ) . Following that, ( 1 ) − ( − 1 ) = 2 and 2 2 = 1
So the simplified version would look something like :
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 ) are also integer solutions.
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I added whole numbers and Thank You!
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Whole numbers includes 0, so (0,1) is still possible.
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It must be known that τ = 2 π , and thus e τ i = e 2 π i . As well as Euler's formula: e x i = c o s ( x ) + i s i n ( x )
Using Euler's formula to compute e 2 π i − e − π 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
Thus p 3 + 2 e 2 π i − e − π i = p 3 + 2 2 = p 3 + 1 The only solution to p 3 + 1 = q 2 where both p and q are integers is p = 2 and q = 3 ,
It follows then that q p = 3 2 = 9