<s>[INST] 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.
 [/INST] 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.
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