# Can $3p^4-3p^2+1$ be square number?

I know $$4p^4+1$$ can't be square number, cause $$4p^4<4p^4+1<4p^4+4p^2+1$$ for all natural number p. But I don't know the way to prove $$3p^4-3p^2+1$$ can(not) be square number. Is there a well known way to prove it?

• $p = 1$ is a square number. – Toby Mak Dec 20 '18 at 11:14
• PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing. – Aqua Dec 20 '18 at 11:33
• You must be fun at parties. – Lucas Henrique Dec 20 '18 at 11:47
• @LucasHenrique Who are you reffering to and what does that have with a math? – Aqua Dec 20 '18 at 12:37

Partial solution if $$p$$ is prime.

Write $$3p^4-3p^2+1=n^2\implies 3p^2(p^2-1) = (n-1)(n+1)$$

If $$p\ne 2$$ (which is not a solution) then $$p^2\mid n-1$$ or $$p^2\mid n+1$$

First case: If $$p^2\mid n-1$$ then $$n+1\mid 3p^2-3$$ so $$n-1= p^2k$$ and $$n+1\leq 3p^2-3$$.

If $$k\geq 3$$ then $$3p^2-3\geq n+1 >n-1 \geq 3p^2$$ which is impossible. So $$k\leq 2$$

$$\bullet$$ If $$k=2$$ then $$n= 2p^2+1$$ so $$2p^2+2\mid 3p^2-3 \implies 2p^2+2\mid 2(3p^2-3)-3(2p^2+2) = -12$$

So $$p^2+1\mid 6 \implies p^2+1\in \{1,2,3,6\}$$ which is impssible.

$$\bullet$$ If $$k=1$$ then $$n= p^2+1$$ so $$p^2+2\mid 3p^2-3 \implies p^2+2\mid (3p^2-3)-3(p^2+2) =-9$$

So $$p^2+2\mid 9 \implies p^2+2\in \{1,3,9\}$$ which is impossible again.

Second case: If $$p^2\mid n+1$$ then $$n-1\mid 3p^2-3$$ so $$n+1= p^2k$$ and $$n-1\leq 3p^2-3$$.

Again, if $$k\geq 3$$ then $$3p^2-3\geq n-1 = n+1-2\geq 3p^2-2$$ which is impossible. So $$k\leq 2$$

$$\bullet$$ If $$k=2$$ then $$n= 2p^2-1$$ so $$2p^2-2\mid 3p^2-3$$ which is impossible.

$$\bullet$$ If $$k=1$$ then $$n= p^2-1$$ so $$p^2-2\mid 3p^2-3 \implies p^2-2\mid (3p^2-3)-3(p^2-2) =3$$

So $$p^2-2\mid 3 \implies p^2-2\in \{-1,1,3\}$$ which is impossible again.

So the answer is negative if $$p$$ is prime.

• what if p is a natural number? Isn't there any way to prove it? – eandpiandi Dec 20 '18 at 12:04

$$3m^2 - 3m + 1 = n^2$$ for any n $$\in$$ Natural, for $$p^2 = m$$.
$$3m^2 - 3m + 1 - n^2 = 0$$ $$\implies m = \frac {3 +\sqrt{12n^2-3}}6$$ $$\implies p^2 = \frac {3 +\sqrt{12n^2-3}}6$$ or $$\implies p^2 = \frac {3 -\sqrt{12n^2-3}}6$$
You can compare it with general formula for Pythagorean triples: $$(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$$, we have: $$(p^2-1)^2+p^2(2p^2-1)$$, here $$k=1$$ and second term must be $$(2p)^2$$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.