Difference of powers pattern

Sit with a table of squares for a while and something strange keeps happening: $8^2-6^2$ comes out to $28$ , and so does $14\times2$ . Try $12^2-9^2$ and you get $63$ , which is exactly $21\times3$ . It is not a coincidence you stumble on once — it happens every single time, for every pair of numbers you can name. That reliability is what turns the identity $(a+b)(a-b)=a^2-b^2$ from a one-off algebra trick into a genuine pattern: something you can tabulate, verify across many rows, and then trust enough to predict a result before you have squared a single number. This article builds that table, pins down the fixed-gap version of the pattern, and uses it to explain a famous "coincidence" — why $99^2-1^2$ , $98^2-2^2$ and $97^2-3^2$ all land suspiciously close together.

The core insight

For any two numbers $a>b$ , the difference of their squares always equals $(a+b)(a-b)$ , so tabulating $a+b$ and $a-b$ alongside $a^2-b^2$ reveals a predictable pattern instead of a coincidence.

The core idea

The identity behind the pattern comes straight from the distributive law: expand $(a+b)(a-b)$ and the cross terms $+ab$ and $-ab$ cancel, leaving only the squares.

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

where: $a$ and $b$ are any two numbers with $a>b$ , $a+b$ is their sum, and $a-b$ is their difference (the "gap" between them). Because this holds for every choice of $a$ and $b$ — not just round or convenient ones — it is safe to build a table and check it row by row, confident that no row can fail:

$a$	$b$	$a^2$	$b^2$	$a^2-b^2$	$(a+b)(a-b)$
5	3	25	9	16	$8\times2=16$
8	6	64	36	28	$14\times2=28$
12	9	144	81	63	$21\times3=63$

Every row matches. The last two columns never disagree, no matter how $a$ and $b$ are chosen — which is exactly why this counts as a pattern worth predicting from, not just an identity worth memorising. A particularly useful special case appears when $a$ and $b$ are always a fixed gap apart — say $a-b=2$ throughout a table, as in the rows above. Then the identity collapses to $a^2-b^2=(a+b)\times2$ : twice the sum. Fix the gap at $5$ instead and the rule becomes $a^2-b^2=5(a+b)$ . Once the gap is known, the entire family of differences-of-squares for that gap can be predicted from a single multiplication, without squaring anything. There is a mirror-image special case too: sometimes it is the sum $a+b$ that stays fixed across a table rather than the gap. In that situation the identity is read the other way around: $a^2-b^2=(a+b)(a-b)$ becomes $a^2-b^2=k(a-b)$ for the fixed sum $k$ . Both shortcuts are the same identity — only the roles of "which factor is constant" and "which factor varies" have swapped. Recognising which one applies to a given table is itself part of the pattern-spotting skill this topic is building: check whether $a+b$ or $a-b$ repeats down the rows, and use that column as your multiplier.

Figure

A three-column strip showing a and b as two bars a fixed gap apart their squared areas and the equal rectangle a plus b times a minus b

The pattern holds for a fixed gap

The goal here is to show, for a whole family of number pairs that share the same gap $a-b$ , that $a^2-b^2$ is not merely equal to $(a+b)(a-b)$ in each isolated case but collapses to a single simple multiple of $a+b$ — turning three separate long calculations into one short rule.

Derivation ADifference of squares for a fixed gap

The goal is to show that once the gap between $a$ and $b$ is fixed at some constant value $k$ , the whole family of differences $a^2-b^2$ reduces to a single multiplication rule, $k(a+b)$ , so that any pair sharing that gap can be checked without computing either square directly.

Step 1. Start from the identity that always holds: $a^2-b^2=(a+b)(a-b)$ . This is true for every pair, regardless of how far apart $a$ and $b$ are.

Step 2. Suppose the gap is fixed: $a-b=k$ for some constant $k$ (for example $k=2$ or $k=5$ , as in the table and the worked example below). Substitute this directly into the identity: $a^2-b^2=(a+b)\times k$ .

Step 3. Read what this means for a table of pairs that all share the same gap $k$ : as $a$ and $b$ climb together while staying $k$ apart, $a^2-b^2$ simply tracks $k$ times their sum. Doubling the sum doubles the difference of squares; there is no need to ever compute $a^2$ or $b^2$ separately once $k$ is known.

Step 4. Check it against the earlier table, where $k=2$ : for $(5,3)$ , $k(a+b)=2\times8=16$ , matching $25-9=16$ . For $(8,6)$ , $2\times14=28$ , matching $64-36=28$ . For $(12,9)$ , $3\times21=63$ — here $k=3$ , not $2$ , since $12-9=3$ , and the formula still holds with the correct gap.

Worked example

Verify the difference-of-squares pattern for three pairs of numbers that are each $5$ apart, and state the single rule connecting $a^2-b^2$ to $a+b$ for that family.

Step 1. Choose three pairs with $a-b=5$ : $(a,b)=(10,5)$ , $(15,10)$ , $(20,15)$ .

Step 2. For $(10,5)$ : $a^2-b^2=100-25=75$ . Check via the identity: $(a+b)(a-b)=15\times5=75$ . Matches.

Step 3. For $(15,10)$ : $a^2-b^2=225-100=125$ . Check: $25\times5=125$ . Matches.

Step 4. For $(20,15)$ : $a^2-b^2=400-225=175$ . Check: $35\times5=175$ . Matches.

Step 5. Since $a-b=5$ is fixed throughout, the identity collapses to $a^2-b^2=5(a+b)$ : the difference of squares is always exactly $5$ times the sum, whenever the two numbers are $5$ apart. Every row confirms it: $75=5\times15$ , $125=5\times25$ , $175=5\times35$ .

Competitive stretch

A student notices $99^2-1^2$ , $98^2-2^2$ and $97^2-3^2$ all look suspiciously related. Here $a+b=100$ stays fixed in every pair instead of the gap. Using the identity, $99^2-1^2=(99+1)(99-1)=100\times98=9800$ ; $98^2-2^2=100\times96=9600$ ; $97^2-3^2=100\times94=9400$ . The three answers are not equal, but each is a multiple of $100$ , falling by $200$ each step — because with $a+b=100$ fixed, $a^2-b^2=100(a-b)$ , and $a-b$ itself falls by $2$ each time ( $98\to96\to94$ ), giving steps of $100\times2=200$ . The results are $9800$ , $9600$ and $9400$ , explained entirely by the fixed-sum version of the pattern.

Common traps

Confusing "difference of squares" with "square of a difference." $a^2-b^2$ is not the same as $(a-b)^2$ ; the two look similar but give different numbers — check with $a=5,b=3$ : $a^2-b^2=16$ but $(a-b)^2=4$ .
Assuming the pattern only works for round, "nice" numbers. The identity $a^2-b^2=(a+b)(a-b)$ is true for every pair without exception, so it is always worth checking rather than dismissing as a fluke of the chosen numbers.
Mixing up which quantity is fixed. A fixed gap ( $a-b=k$ ) gives the rule $a^2-b^2=k(a+b)$ ; a fixed sum ( $a+b=k$ ) gives the different rule $a^2-b^2=k(a-b)$ — spotting which one is constant in a given table decides which shortcut applies.
Building the table without the extra columns. Staring only at raw squared values hides the pattern; adding $a+b$ and $a-b$ as columns makes the match with $a^2-b^2$ obvious immediately.

Ask the tutor: pick your own sequence of number pairs — with a fixed gap or a fixed sum — and ask the AI tutor to help you find and prove which version of the pattern it follows.

How is this different from just using the identity once, like in earlier lessons?

Earlier you probably used $(a+b)(a-b)=a^2-b^2$ to simplify or compute a single pair of numbers. Here the focus shifts to a whole sequence of pairs — a table, or numbers with a shared gap or shared sum — and to noticing that the identity produces a consistent, predictable rule across every row. It is the same algebra, but the skill being tested is pattern recognition: spotting the constant ( $a+b$ , $a-b$ , or their fixed relationship) that makes an entire family of results predictable at once.

Do I always need a=10 type round numbers for the pattern to be useful?

No — the pattern is most useful precisely when $a$ and $b$ are not round. If $a+b$ or $a-b$ happens to be small and easy even though $a$ and $b$ themselves are large or awkward, the identity lets you skip squaring altogether. For example, $997^2-3^2$ is instantly $(1000)(994)=994000$ without ever computing $997^2$ . The pattern works everywhere; it is simply easiest to notice using round numbers first.

What is the quickest way to tell whether a table follows the fixed-gap or the fixed-sum version of the rule?

Compute $a-b$ and $a+b$ for each row and see which one stays the same down the table. If $a-b$ is constant, use $a^2-b^2=k(a+b)$ with $k$ that constant gap. If instead $a+b$ is constant, use $a^2-b^2=k(a-b)$ with $k$ that constant sum. Both are just the identity $(a+b)(a-b)$ with one factor fixed — checking two extra columns tells you instantly which shortcut applies.