AlgebraDifficulty 5-4036 tagged problems

Difference of Squares

Difference of Squares is the habit of using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$a^2 - b^2 = (a - b)(a + b)$$

A difference of two squares always factors into the product of their difference and their sum.

Why It Works

1
Expand the right-hand side: $(a - b)(a + b) = a^2 + ab - ab - b^2$.
2
The middle terms cancel, leaving $a^2 - b^2$ — which is exactly the left-hand side.
3
Run in reverse: spotting $a^2 - b^2$ lets you replace a subtraction with a product, which often exposes roots, divisibility, or cancellation.

Worked Examples

Example 1

Compute $2024^2 - 2023^2$ without a calculator.

1
Recognize a difference of squares with $a = 2024$, $b = 2023$.
2
Factor: $2024^2 - 2023^2 = (2024 - 2023)(2024 + 2023)$.
3
Simplify: $(1)(4047) = 4047$.

Answer:

$4047$

Warm-up

Evaluate $\dfrac{2025^2 - 2024^2}{2025 + 2024}$ without a calculator.

1
The numerator is a difference of squares: $2025^2 - 2024^2 = (2025 - 2024)(2025 + 2024)$.
2
So the fraction is $\dfrac{(2025 - 2024)(2025 + 2024)}{2025 + 2024}$, and the common factor $2025 + 2024$ cancels.
3
What remains is $2025 - 2024 = 1$.

Answer:

$1$

Contest level

Find the sum $1^2 - 2^2 + 3^2 - 4^2 + \cdots + 99^2 - 100^2$.

1
Pair the terms as $(1^2 - 2^2) + (3^2 - 4^2) + \cdots + (99^2 - 100^2)$ — there are $50$ pairs.
2
Each pair is a difference of squares: $(2k-1)^2 - (2k)^2 = \big((2k-1) - 2k\big)\big((2k-1) + 2k\big) = (-1)(4k - 1) = -(4k - 1)$.
3
Sum over $k = 1$ to $50$: $-\sum_{k=1}^{50}(4k - 1) = -\big(4\cdot\tfrac{50\cdot 51}{2} - 50\big) = -(4\cdot 1275 - 50) = -(5100 - 50)$.

Answer:

$-5050$

Olympiad / Challenge

Find all pairs of positive integers $(m, n)$ with $m \ge n$ such that $m^2 - n^2 = 105$.

1
Factor the left side: $m^2 - n^2 = (m - n)(m + n) = 105$.
2
Let $d_1 = m - n$ and $d_2 = m + n$ with $d_1 \le d_2$, $d_1 d_2 = 105$. Since $m = \frac{d_1 + d_2}{2}$ and $n = \frac{d_2 - d_1}{2}$, both $d_1, d_2$ must have the same parity; as $105$ is odd, both are odd automatically.
3
The factor pairs of $105 = 3 \cdot 5 \cdot 7$ with $d_1 \le d_2$ are $(1, 105), (3, 35), (5, 21), (7, 15)$.
4
These give $(m, n) = (53, 52), (19, 16), (13, 8), (11, 4)$, all with $n \ge 1$.

Answer:

$(m, n) \in \{(53, 52), (19, 16), (13, 8), (11, 4)\}$.

Going Deeper

Generalization: over any commutative ring the identity $a^2 - b^2 = (a-b)(a+b)$ holds, and it iterates — $a^4 - b^4 = (a-b)(a+b)(a^2+b^2)$ — which is how repeated difference-of-squares unravels $x^{2^k} - 1$.
Where it appears: it is the engine behind 'mental-math' AMC arithmetic, factoring $n^2 - 1 = (n-1)(n+1)$ in number-theory problems, and representing an integer as a difference of squares (possible iff the integer is odd or a multiple of $4$).
Pitfall: $a^2 + b^2$ does NOT factor over the reals/integers (only over $\mathbb{C}$ as $(a+bi)(a-bi)$). Forcing a 'sum of squares' factorization is a classic error — and when splitting $N = (m-n)(m+n)$, you must discard factor pairs of opposite parity, since $m, n$ would then be non-integers.

Spot the Signal

Look for problems where the key step is using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities.
You can describe the hard part as using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for difference of squares, then rewrite the givens around it.
Name the relation that makes Difference of Squares legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Difference of Squares just because the surface looks familiar; verify the required condition first.
Applying Difference of Squares because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let difference of squares reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities.

Back to techniques

AlgebraDifficulty 5-4036 tagged problems

Difference of Squares

Practice this technique Match my level

The Key Result

$$a^2 - b^2 = (a - b)(a + b)$$

A difference of two squares always factors into the product of their difference and their sum.

Why It Works

1
Expand the right-hand side: $(a - b)(a + b) = a^2 + ab - ab - b^2$.
2
The middle terms cancel, leaving $a^2 - b^2$ — which is exactly the left-hand side.
3
Run in reverse: spotting $a^2 - b^2$ lets you replace a subtraction with a product, which often exposes roots, divisibility, or cancellation.

Worked Examples

Example 1

Compute $2024^2 - 2023^2$ without a calculator.

1
Recognize a difference of squares with $a = 2024$, $b = 2023$.
2
Factor: $2024^2 - 2023^2 = (2024 - 2023)(2024 + 2023)$.
3
Simplify: $(1)(4047) = 4047$.

Answer:

$4047$

Warm-up

Evaluate $\dfrac{2025^2 - 2024^2}{2025 + 2024}$ without a calculator.

1
The numerator is a difference of squares: $2025^2 - 2024^2 = (2025 - 2024)(2025 + 2024)$.
2
So the fraction is $\dfrac{(2025 - 2024)(2025 + 2024)}{2025 + 2024}$, and the common factor $2025 + 2024$ cancels.
3
What remains is $2025 - 2024 = 1$.

Answer:

$1$

Contest level

Find the sum $1^2 - 2^2 + 3^2 - 4^2 + \cdots + 99^2 - 100^2$.

1
Pair the terms as $(1^2 - 2^2) + (3^2 - 4^2) + \cdots + (99^2 - 100^2)$ — there are $50$ pairs.
2
Each pair is a difference of squares: $(2k-1)^2 - (2k)^2 = \big((2k-1) - 2k\big)\big((2k-1) + 2k\big) = (-1)(4k - 1) = -(4k - 1)$.
3
Sum over $k = 1$ to $50$: $-\sum_{k=1}^{50}(4k - 1) = -\big(4\cdot\tfrac{50\cdot 51}{2} - 50\big) = -(4\cdot 1275 - 50) = -(5100 - 50)$.

Answer:

$-5050$

Olympiad / Challenge

Find all pairs of positive integers $(m, n)$ with $m \ge n$ such that $m^2 - n^2 = 105$.

1
Factor the left side: $m^2 - n^2 = (m - n)(m + n) = 105$.
2
Let $d_1 = m - n$ and $d_2 = m + n$ with $d_1 \le d_2$, $d_1 d_2 = 105$. Since $m = \frac{d_1 + d_2}{2}$ and $n = \frac{d_2 - d_1}{2}$, both $d_1, d_2$ must have the same parity; as $105$ is odd, both are odd automatically.
3
The factor pairs of $105 = 3 \cdot 5 \cdot 7$ with $d_1 \le d_2$ are $(1, 105), (3, 35), (5, 21), (7, 15)$.
4
These give $(m, n) = (53, 52), (19, 16), (13, 8), (11, 4)$, all with $n \ge 1$.

Answer:

$(m, n) \in \{(53, 52), (19, 16), (13, 8), (11, 4)\}$.

Going Deeper

Generalization: over any commutative ring the identity $a^2 - b^2 = (a-b)(a+b)$ holds, and it iterates — $a^4 - b^4 = (a-b)(a+b)(a^2+b^2)$ — which is how repeated difference-of-squares unravels $x^{2^k} - 1$.
Where it appears: it is the engine behind 'mental-math' AMC arithmetic, factoring $n^2 - 1 = (n-1)(n+1)$ in number-theory problems, and representing an integer as a difference of squares (possible iff the integer is odd or a multiple of $4$).
Pitfall: $a^2 + b^2$ does NOT factor over the reals/integers (only over $\mathbb{C}$ as $(a+bi)(a-bi)$). Forcing a 'sum of squares' factorization is a classic error — and when splitting $N = (m-n)(m+n)$, you must discard factor pairs of opposite parity, since $m, n$ would then be non-integers.

Spot the Signal

Look for problems where the key step is using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities.
You can describe the hard part as using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for difference of squares, then rewrite the givens around it.
Name the relation that makes Difference of Squares legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Difference of Squares just because the surface looks familiar; verify the required condition first.
Applying Difference of Squares because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let difference of squares reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using $a^2-b^2=(a-b)(a+b)$ to simplify expressions, integer equations, and geometric quantities.