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once again a question on if my proof is correct, since the way it is written differs much with how the book proofs it. I know that the $n$ chosen does not make much of a difference in this case, I am mostly doubting if the steps taken towards the conclusion are enough to prove $2n^2 - 5n + 2$ is prime

The Question given is :
"$\exists n \in \mathbb{Z}$ such that $2n^2 - 5n + 2$ is prime"

Here's the proof I have written :
Suppose an integer $n$
Let $n = 0$
Then by substitution $2n^2 - 5n + 2 = 2(0)^2-5(0)+2$
$=0-0+2$
$=2$, which is a prime number
$\therefore \exists n \in \mathbb{Z}$ such that $2n^2 - 5n + 2$ is prime, where $n=0$

This is the answer given in the solution book :

When $n = 3$,
Now put the value of $n = 3$ in equation $2n^2 - 5n + 2$,
This implies;
$2n^2 - 5n + 2 = 2(3)^2-5(3)+2$
$= 2(9)-15+2$
$= 18-13$
$= 5$
This implies;
$2n^2 - 5n + 2 = 5$
Here, $5$ is a prime number.
So,$2n^2 - 5n + 2$ is also prime.
Hence proved, there is an integer $n$ such that $2n^2 - 5n + 2$ is prime.

Diceble
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  • One integer giving a prime is enough. Proof by inspection is the only way in such cases. – Peter Jan 17 '23 at 17:08
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    Yes your proof is correct. Are you sure the question does not ask for a positive integer? – Vivaan Daga Jan 17 '23 at 17:09
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    $(x-2)(2x-1)$ helps I guess –  Jan 17 '23 at 17:13
  • @Shinrin-Yoku Yes, I mostly questioned myself due to the fact they have more steps in between the beginning and the end. I thought I maybe lacked some steps in my proof. – Diceble Jan 17 '23 at 17:20
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    The solutions are way too long. With $f(n)=2n^2-5n+2=(n-2)(2n-1)$ it is enough to say that $f(3)=5$ is prime, or $f(0)=2$ is prime, if $n=0$ is allowed. For all other $n$ the integer $f(n)$ is composite. Compare this with this question. – Dietrich Burde Jan 17 '23 at 17:21
  • In short, your proof is cumbersome, but correct , unless a positive integer is explicitely demanded. – Peter Jan 17 '23 at 17:29
  • @DietrichBurde Thank you for your feedback. I lack the ability to see what you are trying to show me with comparing this question, with the question you mentioned. Could you explain this in more detail? – Diceble Jan 17 '23 at 17:35
  • @DietrichBurde $f(-1)=-1,$ which I would say is neither prime nor composite, but of course that still leaves $n=0$ and $n=3$ as the only witnesses of the conclusion. – David K Jan 17 '23 at 17:36
  • @Peter Thank you for your feedback. – Diceble Jan 17 '23 at 17:36
  • Any feedback on the -1, then I can take this into account into future questions. – Diceble Jan 17 '23 at 17:37
  • @Diceble Usually, only integers larger than $1$ are considered to be "composite" or "prime". We could extend this to negative integers smaller than $-1$ which is sometimes done. I have however never heard of "composite" or "prime" in context with $-1$ , $0$ or $1$. It is reasonable to consider those numbers to be neither because neither the units $-1$ and $1$ nor $0$ are prime elements in the ring $\mathbb Z$. – Peter Jan 17 '23 at 17:42
  • I don't know why the question was downvoted. I would call the proofs "painstaking" rather than cumbersome. A teacher might encourage such simple proofs to have this many steps when the student is new to writing proofs. When you have more experience you might just write, "When $n=0,$ $2n^2 - 5n + 2=2,$ which is prime, therefore there is an integer $n$ such that $2n^2 - 5n + 2$ is prime." For the right audience you might even stop at "which is prime". – David K Jan 17 '23 at 17:44
  • @Peter Thank you for your further explanation – Diceble Jan 17 '23 at 17:49
  • @DavidK I am indeed new to writing proofs, Thank you for showing another way which is a more experienced way of writing the same proof. – Diceble Jan 17 '23 at 17:50
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    @Diceble Do not care about strange downvotes. I receive them often without an apparent reason or an explanation why it was downvoted. – Peter Jan 17 '23 at 17:51

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