proof that there doesn't exist integers $a, b, c>0$, that satisfy $a^2bc+2$, $ab^2c+2$ and $abc^2+2$ are all square numbers. please proof that there doesn't exist a,b,c which  are all positive integers, that satisfy a^2bc+2, ab^2c+2 and abc^2+2 are all square numbers.
I know squares remain 1 or 0 when divided by 4, so a^2bc, ab^2c and abc^2 must remain 2 or 3 when divided by 4. Since a^2,b^2,c^2 must remain 1 ,we can proof ab,bc,ac can’t all remain 2 or 3 at the same time.If any two of a,b,c remain the same, one of ab, bc or ac will remain 1.Hence a,b,c must remain 1,2 and 3.
But this is not a contradiction to my suppose. Which part of my proof is wrong, or should I use other way to solve this problem?
I’ve tried mod 3 and it doesn’t work either.
Any help is appreciated.
 A: You are almost there.  If $a,b,c$ are $1,2,3 \bmod 4$ we might as well say $a \equiv 1, b\equiv 2, c \equiv 3 \bmod 4$.  Then $ab^2c \equiv 0, ab^2c+2 \equiv 2 \bmod 4$ so it cannot be a square.
A: Here is a different way to prove it. Assume you have them all being square numbers, i.e., there are integers $x$, $y$ and $z$ such that
$$a^2bc + 2 = x^2 \tag{1}\label{eq1A}$$
$$ab^2c + 2 = y^2 \tag{2}\label{eq2A}$$
$$abc^2 + 2 = z^2 \tag{3}\label{eq3A}$$
First, assume all $a$, $b$ and $c$ are odd. Then so must be $x$, $y$ and $z$. Thus, modulo $4$, since all odd squares are congruent to $1$ modulo $4$, we get by moving the $2$ to the right (since $1 - 2 \equiv 3 \pmod 4$) that
$$bc \equiv 3 \pmod 4 \tag{4}\label{eq4A}$$
$$ac \equiv 3 \pmod 4 \tag{5}\label{eq5A}$$
$$ab \equiv 3 \pmod 4 \tag{6}\label{eq6A}$$
Subtracting \eqref{eq5A} from \eqref{eq4A} gives
$$c(b - a) \equiv 0 \pmod 4 \implies b \equiv a \pmod 4 \tag{7}\label{eq7A}$$
However, in \eqref{eq6A}, this gives $a^2 \equiv 3 \pmod 4$. Thus, this is not possible. As such, at least one value among $a$, $b$ and $c$ must be even. Let this be, WLOG, $a$, so $a^2bc \equiv 0 \pmod 4$. However, in \eqref{eq1A}, the LHS is congruent to $2$ modulo $4$, which is not allowed for square numbers.
This shows it's not possible to do what is requested.
A: You idea will work.  We just need to be careful.
Suppose all $3$ $a^2bc + 2$ and $ab^2c + 2$ and $abc^2+2$ are square numbers.
Then all $a^2bc, ab^2c, abc^2$ will be $\equiv 2,3 \pmod 4$.
Now suppose $a$ is even.  Then $a^2 \equiv 0\pmod 4$.  And $a^2bc \equiv 0 \pmod 4$ . That is a contradiction.  So $a$ is odd.
The same argument will prove that $b$ and $c$ are odd and thus either $1,3 \pmod 4$.
So $a^2bc, ab^2c, abc^2$ are all odd and $\equiv 3\pmod 4$.
$a^2,b^2,c^2$ are all $\equiv 1\pmod 4$, so $bc,ac, ab$ are all $\equiv 3\pmod 4$.
If $b,c$ are both $\equiv 1\pmod 4$ or both $\equiv 3 \pmod 4$ then $bc\equiv 1\pmod 4$.  So $b,c$ must have different equivalencies.
And the same argument will show that $a$ and $c$, and $a$ and $b$ will have different equivalencies from each other.
So $b$ and $c$ both have a different equivalency than $a$.  But as there are only two options, $1$ or $2\pmod 4$ that would mean $b$ and $c$ must have the same as each other.
Which is a contradiction.
It's possible for two to be perfect squares but not all three.
