1. Key Concepts and Formulas
   • Difference of Squares: $a^2 - b^2 = (a-b)(a+b)$. This is fundamental for simplifying pairs of terms in the series.
   • Sum of an Arithmetic Series: The sum of the first $N$ natural numbers is $1+2+3+\ldots+N = \frac{N(N+1)}{2}$. This will be used to sum the simplified terms.
   • Prime Factorization: Decomposing a number into its prime factors is crucial for identifying $m$ and $n$ from the equation $m^2 n = \text{constant}$.
   • Greatest Common Divisor (GCD): The condition $\operatorname{gcd}(m, n)=1$ ensures a unique solution for $m$ and $n$.

---

1. Step-by-Step Solution

Step 1: Analyze and Group the Series Terms We are given the series $S = 1^2-2^2+3^2-4^2+\ldots +(2021)^2-(2022)^2+(2023)^2$. The series has $2023$ terms. We can group consecutive terms into pairs of the form $(2k-1)^2 - (2k)^2$. The last term, $(2023)^2$, will remain unpaired. $S = (1^2-2^2) + (3^2-4^2) + \ldots + (2021^2-2022^2) + (2023)^2$

Step 2: Apply the Difference of Squares Formula Using the formula $a^2 - b^2 = (a-b)(a+b)$ for each pair:

• $1^2-2^2 = (1-2)(1+2) = (-1)(3)$
• $3^2-4^2 = (3-4)(3+4) = (-1)(7)$
• ...
• $2021^2-2022^2 = (2021-2022)(2021+2022) = (-1)(4043)$

Substituting these back into the series: $S = (-1)(1+2) + (-1)(3+4) + \ldots + (-1)(2021+2022) + (2023)^2$

Step 3: Simplify the Series Sum Factor out $-1$ from the paired terms: $S = -[(1+2) + (3+4) + \ldots + (2021+2022)] + (2023)^2$ The terms inside the bracket form the sum of the first $2022$ natural numbers: $1+2+3+\ldots+2022$. $S = -\left(\sum_{i=1}^{2022} i\right) + (2023)^2$

Step 4: Calculate the Sum of the Arithmetic Series Using the formula for the sum of the first $N$ natural numbers, $\frac{N(N+1)}{2}$, with $N=2022$: $\sum_{i=1}^{2022} i = \frac{2022(2022+1)}{2} = \frac{2022 \times 2023}{2} = 1011 \times 2023$ Substitute this back into the expression for $S$: $S = -(1011 \times 2023) + (2023)^2$

Step 5: Factor and Simplify the Expression for S We can factor out $2023$ from the expression: $S = 2023(2023 - 1011)$ $S = 2023(1012)$

Step 6: Equate the Series Sum with the Given Equation We are given that $S = 1012 m^2 n$. Equating our calculated sum with this: $2023 \times 1012 = 1012 m^2 n$ Divide both sides by $1012$ (since $1012 \neq 0$): $m^2 n = 2023$

Step 7: Find m and n using Prime Factorization and GCD Condition We need to find integers $m$ and $n$ such that $m^2 n = 2023$ and $\operatorname{gcd}(m, n)=1$. First, find the prime factorization of $2023$: $2023 = 7 \times 289 = 7 \times 17^2$. So, $m^2 n = 17^2 \times 7$. Since $m^2$ must be a perfect square, we can identify $m^2 = 17^2$ and $n=7$. This gives $m=17$ (taking the positive root) and $n=7$. Now, check the GCD condition: $\operatorname{gcd}(17, 7)$. Since 17 and 7 are distinct prime numbers, their GCD is 1. This condition is satisfied.

Step 8: Calculate $m^2 - n^2$ Finally, we need to compute $m^2 - n^2$: $m^2 - n^2 = 17^2 - 7^2$ $m^2 - n^2 = 289 - 49$ $m^2 - n^2 = 240$

1. Common Mistakes & Tips

   • Sign Errors: Be meticulous with the negative signs when applying the difference of squares formula and when factoring. A mistake here can lead to an incorrect sum.
   • Unpaired Term: Ensure the last term of the series is handled correctly and not accidentally paired.
   • GCD Verification: Always verify the $\operatorname{gcd}(m, n)=1$ condition. It's crucial for uniquely determining $m$ and $n$ from their prime factorizations. For example, if $m^2 n = 72 = 36 \times 2$, then $m=6, n=2$ would be invalid because $\operatorname{gcd}(6,2) \neq 1$.

2. Summary The problem involved evaluating an alternating sum of squares. By grouping terms and applying the difference of squares formula, the series was simplified into a form involving the sum of an arithmetic progression. After calculating the sum and equating it to the given expression $1012 m^2 n$, we used the prime factorization of $2023$ and the $\operatorname{gcd}(m, n)=1$ condition to uniquely determine $m=17$ and $n=7$. Finally, $m^2 - n^2$ was calculated.

The final answer is $\boxed{240}$.