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I have a general idea of how to prove this statement, but I'm having trouble translating the statement to predicate logic. My attempt is as follows:

$P(x): x$ is prime

$\forall m \in \mathbb{N}, \exists x \in \mathbb{N}, x = m^3 -1 \land P(x) \Rightarrow x = 7 $

Adrian Keister
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Joel Emiid
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  • You should change $\exists x\in\mathbb{N}$ to $\forall x\in\mathbb{N},$ and then I think it's totally fine. – Adrian Keister Nov 26 '19 at 18:12
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    $x$ seems unnecessary...$\forall m\in \mathbb N$, $P(m^3-1)\implies m=2$ or some simple variant of that. – lulu Nov 26 '19 at 18:15
  • Would I be able to let $x$ be equal to a number if it's universally quantified though? @AdrianKeister – Joel Emiid Nov 26 '19 at 18:15
  • Sure! You're not "letting" $x$ be equal to a number: you're saying that, for any number $x,$ if it satisfies certain properties, it must be $7.$ – Adrian Keister Nov 26 '19 at 18:20

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Adrian is right that the $\exists$ should be a $\forall$. Why? Because otherwise, we can make the conditional vacuously true simply by pointing to any $x$ for which the antecedent is false, and thus the statement would be true without saying anything interesting about any numbers that do have the property of being one less than a perfect cube. Indeed, note what I used there: 'any'! That is, it is true that for any number whose successor is a perfect cube, that number will have to be $7$, and so we really need to use a universal quantifier, i.e.:

$\forall m \in \mathbb{N}, \forall x \in \mathbb{N}, ((x = m^3 -1 \land P(x)) \Rightarrow x = 7)$

or, what is the equivalent:

$\forall x \in \mathbb{N}, (\exists m \in \mathbb{N}, (x = m^3 -1 \land P(x)) \Rightarrow x = 7)$

I am wondering if maybe you used $\exists \ x$ because when translating 'there is only ...' or 'there is a unique ...' you are typically not told what this special something is, and hence you do need to introduce an existential. That is, if the sentence was: 'There is only one number whose successor is a perfect cube', you'd do:

$\exists y \in \mathbb{N}, \forall x \in \mathbb{N}, (\exists m \in \mathbb{N}, (x = m^3 -1 \land P(x)) \Rightarrow x = y)$

But in this case we know what this 'special something' is: it is $7$, and so we don't need the $\exists$. But, we do need the $\forall$ in order to rule out any numbers other than $7$ to have the property in question.

However, notice that we actually want to do something more than ruling out anything other than $7$ to have this property. You also want to say that $7$ actually has this property. That is, the way the statement is now, we don;t know whether $7$ has the property or not, but the statement clearly suggests that it does. So, I would do:

$\forall x \in \mathbb{N}, (\exists m \in \mathbb{N}, (x = m^3 -1 \land P(x)) \Leftrightarrow x = 7)$

Bram28
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  • Wouldn't an implication work as well? In a proof of this statement I saw (https://math.stackexchange.com/questions/671076/is-7-the-only-prime-followed-by-a-cube), I'm assuming a direct implication was used instead of a double implication like you used to show that 7 is only the number with this property – Joel Emiid Nov 26 '19 at 18:32
  • @JoelEmiid All those proofs do indeed only show the one-sided implication: the only way for a number to possible have the property that its successor is a perfect cube is for that number to be $7$. But, that does not immediately show that $7$ actually has that property. Again, it only rules out numbers other than $7$ to have that property, because for any numbers other than $7$ you run into a contradiction. Look at the Accepted Answer: it never points out that you do get a solution for $n=2$, and hence that $n^3-1=7$ is an actual solution. – Bram28 Nov 26 '19 at 18:44
  • @JoelEmiid So, you still need to verify that $7$ actually has the property in order to say that $7$ is 'the lone' one, because otherwise it leaves the possibility that no number has this property. That is, if no number has that property, then you would certainly not say that $7$ is the lone number with that property. – Bram28 Nov 26 '19 at 18:45