Consecutive Perfect Squares

Our solution will be of the form

$n^2 + (n+1)^2 + (n+2)^2 = (n+a)^2 + (n+a+1)^2$, where $a > 2$.

Expanded, this equals $n^2 + (4 - 4a)n + (4 - 2a - 2a^2)$, which is quadratic in $n$. Since we require $n$ to be an integer, it only makes sense to put this into the quadratic formula and find when our discriminant is equal to $0$.

$n = \frac{-(4-4a) \pm \sqrt{(4-4a)^2 - 4(4 - 2a - 2a^2)}}{2}$

Expanded and reduced, we come to

$n = 2a - 2 \pm \sqrt{6(a-1)(a)}$

Thus, $6(a-1)(a)$ is a perfect square.

When $a = 3$, we have the condition given in the premise of the challenge.

Since $gcd(a-1, a) = 1$, it must be that one of the factors is equal to $6 * m^2$ and the other is equal to $n^2$.

If $m = 1$, then the other number is $5$ or $7$, so that doesn't work. Trying $m = 2$, we find a solution in $a = 25$.

Going back to our original equation of $n^2 + (n+1)^2 + (n+2)^2 = (n+a)^2 + (n+a+1)^2$, we substitute $a = 25$ and expand to get $n^2 - 96n - 1296 = (n - 108)(n + 12) = 0$, thus $n = 108$.

Our desired number is $133^2 + 134^2 = 35645$.

We wish to find $x,y,z \in \mathbb{N}$ such that $x = y^2 + (y + 1)^2 = (z - 1)^2 + z^2 + (z + 1)^2$ Rewrite the second equality as $(2y + 1)^2 - 6z^2 = 3$ This is generalized Pell's equation that can be solved as follows, note the fundamental solution $(2y + 1,z) = (3,1)$ and consider the related Pell's equation $a^2 - 6b^2 = 1$ with fundamental solution $(5,2)$. Note that norm $N$ in $\mathbb{Z}\left[\sqrt{6}\right]$ is multiplicative, hence $N\Big(\left(5 + 2\sqrt{6}\right)^n\cdot \left(3 + \sqrt{6}\right)\Big) = 3$ That means $2y + 1 + z\sqrt{6} = \left(5 + 2\sqrt{6}\right)^n\cdot \left(3 + \sqrt{6}\right)$ Plugging in all $n \in \mathbb{N} \cup \{0\}$ generates infinite family of solutions (I think it's been proven that it also generates all solutions, but I am not sure), but we only care about the first few solutions, so we get $(2y + 1,z) \in \{ (3,1), (27,11), (267,109),\dots\}$ The solution $(27,11)$ corresponds to $x = 365$, hence consider the solution $(267,109)$, which gives us $x = 35645 = 133^2 + 134^2 = 108^2 + 109^2 + 110^2$ and the answer is $\boxed{645}$.

We make two observations:

Modulo 3: the sum of three consecutive squares, is always 2, and the only solution for the sum of two consecutive squares to be 2 is when they are equivalent to 1^2 and 2^2 Modulo 2: the sum of two consecutive squares is always odd, so the middle one of the triple must be odd.

So any solution can be written with integers a and b, in the form

$(2a)^2+(2a+1)^2+(2a+2)^2= (3b+1)^2+(3b+2)^2$

This equation is equivalent to $2a(a+1)=3b(b+1)$ The left hand side of his expression is divisible by 4, while the right hand side is divisible by 3, so $a \in \{0,2\} \pmod 3 \text{ and } b \in \{0,3\} \pmod 4$

Furthermore, (mod 5) the left hand side can only take values 0, 2 or 4 while the right hand side only takes the values 0, 1 or 3. They can only be equal if the expression is a multiple of 5. Hence $a,b \in\{0,4\} \pmod 5$

Combined $a \in \{0,5,9,14\} \pmod {15}$ $b \in \{0,4,15,19\} \pmod {20}$

Also, writing $2a^2+2a-3b(b+1)=0$ we have $a=\frac{-2 \pm \sqrt{4+24b(b-1)}}{4}$ or simpler $a=\frac{\sqrt{1+6b(b+1)}-1}{2}$, so $6b(b+1)+1\text { must be an (odd) square}$

Time to try a few values:

$b=0, a=0: 1^2+2^2=0^2+1^2+2^2$. Correct, but uninteresting and too small.

$b=4: 6 \cdot 4 \cdot (4+1)+1=121=11^2, a=5, n=365$.

$b=15: 1441$ is not a square, so no integer value for a satisfies this.

Also for $b \in \{19, 20, 24, 35, 39, 40\}$ , $6b(b+1)+1$ is not a square.

But for $b=44$ it is: $b=44, 6 \cdot 44 \cdot (44+1)+1=11881=109^2$ bingo! $a=54, 2a=108, 3b +1=133$ and $108^2+109^2+110^2=133^2+134^2=\boxed{35645}$.

Let $m,n, m<n$ such that:

$m^2+(m+1)^2+(m+2)^2=(n+1)^2+(n+2)^2$

We get this equation:

$3m^2+6m=2n^2+6n$,

From this equation $m$ must be even.

We have condition

$n \equiv 0 \pmod{6}$,

Check this condition , the first numbe after $n=12$ satisfy this condition is $n=132$,

where $m=108$

Hence, the last three digit for next number $133^2+134^2 \equiv \fbox{645} \pmod{1000}$