Perfect Squares

The six pairwise products are:

$(1).\; \; a\left ( a+2 \right )$

$(2).\; \; b\left ( b+2 \right )$

$(3).\; \; \left ( a+2 \right )\left ( b+2 \right )$

$(4).\; \; a\cdot b$

$(5).\; \; a\left ( b+2 \right )$

$(6).\; \; b\left ( a+2 \right ).$

First of all, (1) and (2) cannot be perfect squares, because

$a\left ( a+2 \right )=a^{2}+2a=(a+1)^{2}-1,$ and

$b\left ( b+2 \right )=b^{2}+2b=(b+1)^{2}-1.$

So now, the upper bound of the maximum value of $S$ is 4, as (1) and (2) are eliminated. Let us assume the maximum value of $S$ is indeed 4 (i.e. (3), (4), (5) and (6) can each form a perfect square simultaneously) --

Consider (3) and (4), if they satisfy the assumption mentioned above, then

$a\cdot b\cdot a(b+2)$ is also a perfect square (the product of two perfect squares is obviously another perfect square, the proof of this is omitted), and

$a\cdot b\cdot a(b+2)=a^{2}b(b+2)$ , which means that $b(b+2)$ is a perfect square, but this contradicts with the previous proof that $b(b+2)$ cannot be a perfect square. Therefore the maximum value of $S$ is less than 4.

We then assume the maximum value of $S$ is 3:

There are 2 possible matchings: (3), (4), (6) and (3), (5), (6), since (4) and (5) do not match with each other.

If the above assumption is true, let us consider (3), (6):

$(a+2)(b+2)\cdot b(a+2)=b(b+2)\cdot (a+2)^{2}$ , which means that $b(b+2)$ is a perfect square. (Using the same technique we can disprove the matching (3), (5) and (4),(6).) As the two possible matchings (3), (4), (6) and (3), (5), (6) both contain (3) and (6), the assumption is not true ( this is similar to the previous proof by contradiction about (4) and (5)).

We then reduce the maximum value of $S$ to 2:

Choose matchings that does not contain (1) or (2), and is not (3), (6) nor (4), (5) nor (3), (5) nor (4), (6),

we get (3), (4) and (5), (6), which are the same after rearranging, i.e. $ab(a+2)(b+2)$ . This does not contradict with any of the previous statements, and it is possible to let $a\cdot b$ and $(a+2)(b+2)$ both be perfect squares simultaneously, with a and b not equal to each other, for example, $a\cdot b=5^{2}$ and $(a+2)(b+2)=9^{2}$ with $\left\{\begin{matrix} a=1 & & \\ b=25& & \end{matrix}\right.$ .

Thus, the maximum value for $S$ is 2.

This solution is not mine: We show the maximum is $2$ . Clearly, suppose $(a_0,b_0)$ are two solutions of the Pell's equation $4x^2-y^2=3$ . Now $a=8(a_0^2-1),b=2(a_0^2-1)$ gives $ab=(4(a_0^2-1))^2, (a+2)(b+2)=(8a_0^2-6)(2a_0^2)=(2a_0b_0)^2$ . So (2) is achieved. 6 or 5 is impossible: Let $a(a+2),b(b+2)$ none of them are squares since they are one less than a square. Thus we cannot have them being a natural square. 4 is impossible: $ab=m^2,a(b+2)=n^2,(a+2)b=p^2,(a+2)(b+2)=q^2\implies mq=pn$ also see that $p^2+n^2+4=m^2+q^2\implies (m-n)^2=(p-q)^2+4(\text{since }mq=pn)\implies p=q$ . But then $a=b$ contradicting distinctness. 3 is impossible: For this to happen one of these pairs must be simultaneously squares: $\{(ab,a(b+2)),(ab,(a+2)b),(a+2)(b+2),a(b+2)),((a+2)(b+2),(a+2)b)\}$ but multiply any two of them and since their product is a perfect square then $a^2+2a$ or $b^2+2b$ must be a perfect square as well. Which is impossible. Hence $\text{max}(S)\le 2$ .