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Prove or Disprove that there exists an integer $n$ such that $4n^2 -12n +8$ is prime.

So we have a strictly positive and even term minus a strictly even term plus an even. Hmm. A counterexample maybe, but I'd rather not guess a number to prove/disprove it. Which I doubt is as easy it sounds.

Since it looks to be strictly even it would mean it can't be prime since integers greater than 2 that are even are not prime?

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    $\begingroup$ What the heck? Factor out a 2? $\endgroup$
    – Randall
    Sep 2, 2017 at 13:48
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    $\begingroup$ Better to factor out a $4$ as this removes the prime $2$ as well as the odd ones. $\endgroup$
    – lulu
    Sep 2, 2017 at 13:49
  • $\begingroup$ @lulu LOL, touche. $\endgroup$
    – Randall
    Sep 2, 2017 at 13:49
  • $\begingroup$ If you factor out a 2, you have $2n^2-6n+4=1$ as your only chance to find $n$... $\endgroup$
    – Eleven-Eleven
    Sep 2, 2017 at 13:50
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    $\begingroup$ Question definitely needs more answers. $\endgroup$
    – Randall
    Sep 2, 2017 at 13:51

5 Answers 5

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$4n^2-12n+8=4(n^2-3n+2)$ can't be prime because it is a multiple of $4$

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  • $\begingroup$ Technically, there can be a prime that is a multiple of $2$, namely $2$ itself, so it is better to use the argument with $4$. $\endgroup$
    – DWe1
    Sep 2, 2017 at 13:51
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    $\begingroup$ @DWe1 Thanks, corrected! $\endgroup$
    – coconut
    Sep 2, 2017 at 13:51
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Since $4$ divides $4n^2 - 12n + 8$, its prime factorization already has the factors $2^2$. There exists no number that is prime and has more than one prime factor. Hence, there cannot be a prime number of this form.

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  • $\begingroup$ You are factually correct, but you are not contradicting anything I said. $0$ is not a prime number. $\endgroup$
    – DWe1
    Sep 2, 2017 at 15:15
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note that $$4n^2-12n+8=4(n^2-3n+2)$$

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A lot of this website's regulars looked at that question and thought "Come on, that's way too easy." As you learn more, you will look back on this and then it will look a lot easier to you then than it does now.

At the same time you will also run into expressions for which it's not as obvious that they don't generate primes. I will use $f(n) = 4n^2 - 12n + 8$ to show you a couple of things you can do to find your way with the more difficult expressions.

Notice that there is only one variable, $n$, which is likely an integer, presumably positive. Still, I would try computing a few values, $-5 < n < 5$. This gives me: 168, 120, 80, 48, 24, 8, 0, 0, 8, 24, 48. It looks like $f(n)$ is never negative, even when $n$ is, and specifically, if $n < 0$ then $f(n) = f(|n| + 3)$.

Already this is enough to suggest that all $f(n)$ are divisible by 8. Try $$\frac{4n^2 - 12n + 8}{8} = \frac{n^2}{2} - \frac{3n}{2} + 1.$$ Not quite sure what to make of that, so let's try dividing by 4 instead: $$\frac{4n^2 - 12n + 8}{4} = n^2 - 3n + 2.$$

If you compute $n^2 - 3n + 2$ for a few small values of $n$, you will see that this expression also gives even numbers, though only some of them are also divisible by 4: 0, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, etc.

That sequence of integers looks familiar. Go to Sloane's OEIS and put it in: you won't get any results. Try dropping the first two zeroes and you'll get http://oeis.org/A002378 as the first result. These are the "oblong" numbers, $n^2 + n$, which are twice the triangular numbers.

In fact, if you search for $4n^2 - 12n + 8$ using the range $3 < n < 10$, you should get http://oeis.org/A033996 as the first result. That's the triangular numbers times 8.

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$$2 \mid (4n^2-12n+8) \ \ \ \ \ \ \text{and} \ \ \ \ \ \ 4 \mid (4n^2-12n+8).$$

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