I Love Squares

Relevant wiki: Pell's Equation

Let $A,B,C$ take the values of $a,a+1, a+2$ , respectively, and
let $P,Q,R,S$ take the values of $b,b+1,b+2, b+3$ , respectively,
where $a$ and $b$ are non-negative integers.

Substituting these values into the equation $A^2+B^2+C^2= P^2+Q^2+R^2+S^2$ and simplifying gives

$3a^2 + 6a = 4b^2 +12b + 9 \quad \Leftrightarrow\quad 3(a+1)^2 - 3 = (2b+3)^2 \; .$

Since LHS is divisible by 3, then so is RHS, so $3 | (2b+3) \Rightarrow 3 | b$ . Let $b = 3c$ , where $c$ is also a non-negative integer,

$3(a+1)^2 - 3 = (6b+3)^2 \quad \Leftrightarrow \quad (a+1)^2 - 1 = 3(2c+3)^2 \quad \Leftrightarrow \quad Y^2 - 3Z^2 = 1 \; ,$

where $Y = a+1, Z = 2c+3$ . This final equation represents a Pell's equation .

A fundamental solution to this equation is $(Y_1 , Z_1) = (2,1)$ . And the family of solutions to this diophantine equation satisfies

$\begin{cases} Y_k =Y_{k-1} Y_1 + n Z_{k-1} Z_1 \\ Z_k = Y_{K-1} Z_1 + Z_{K-1} Y_1 \; . \end{cases}$

With this recurrence relation, we can obtain $(Y,Z) = (2,1), (7,4),(26,15),(97,56),(362,209), (1351,780), (5042, 2911), \ldots$ .

Knowing that $Y = a+1, Z = 2c +3$ , and nothing that $c$ is a non-negative integer, then some of the solutions above are not fulfilled. We are left with

$(a,c) = (1,0), (25,63), (361,936), (5041, 13095), \ldots \quad \Rightarrow \quad (a,b) = (1,0), (25,21), (361,312), (5041, 4365) , \ldots .$

Trial and error shows that only the first $\boxed 3$ solutions satisfy the constraint that $0< X = A^2+B^2+C^2= P^2+Q^2+R^2+S^2 < 10^6$ .