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

Question

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?

Answer 1

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!