# Solve equation $n^4+n^2+1=p$, where p is prime number

If $$n \in \mathbb N$$ and $$p$$ is prime number solve equation $$n^4+n^2+1=p$$ i can write that like this $$n^4+2n^2-n^2+1=p$$ $$(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)=p=1 \cdot p$$ Since $$n^2+1+n>1$$ then $$n^2+1-n=1$$ so $$n=0$$ or $$n=1$$ if I put $$n=0$$ then $$p=1$$ that is not true since one is not prime number, then $$n=1$$ and $$p=3$$, is this ok?

• Yes. ${}{}{}{}{}$ – Jean-Claude Arbaut Nov 21 '18 at 20:09
• ah it depends on what level of seriousness you have in number theory really, for stack exchange yes by the looks of the popular vote is saying this proof is ok, but this would not qualify by my standards. – Adam Nov 25 '18 at 3:27
• You have stated that your aim is to solve an equation when one does this, it means they have obtained all possible solutions to that equation given it's stated definitions. If $p$ can be any prime, you have a solution set that is infinite right? You have stated nothing of the nature of the cardinality of your solution set and so I do recommend you looking into learning about this. – Adam Nov 25 '18 at 3:30

## 1 Answer

Besides punctuation, which I'd use more of in your case, the proof is perfect. Perhaps elaborate on some things a little more (like emphasizing you're taking $$n^4+2n^2+1$$ and turning those terms specifically into $$(n^2+1)^2$$, or showing that $$n^2+1+n>1$$) but the level of detail and preciseness always depends on the context.

In short, nicely done. Good work!