A Formula That Shows Up Wherever Distances Do
Every time you measure the diagonal of a rectangle, the hypotenuse of a right triangle, or the magnitude of a 2D vector, this identity is doing the work.
The a square plus b square formula rewrites a sum of two squares in two equivalent ways:
$$a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab.$$
Both forms are useful — pick whichever matches the data you have.
The Formula
For any real numbers $a$ and $b$:
$$\boxed{;a^2 + b^2 = (a + b)^2 - 2ab;}$$
Equivalently:
$$a^2 + b^2 = (a - b)^2 + 2ab.$$
Quick facts.
Type: algebraic identity (true for all real $a, b$).
Reads as: "sum of squares equals the square of the sum minus twice the product."
Grade introduced: CCSS-M HSA-SSE.A.2; NCERT Class 8 Chapter 9 — Algebraic Expressions and Identities.
Geometric meaning: when $a, b$ are the legs of a right triangle, $a^2 + b^2 = c^2$ — the Pythagorean theorem.
Does not factor over the reals. Unlike $a^2 - b^2$, the sum of two real squares has no real factoring (over the complex numbers, $a^2 + b^2 = (a + bi)(a - bi)$).
Two Quick Proofs
Proof 1 — Subtract from $(a+b)^2$.
Start with the binomial-square identity:
$$(a + b)^2 = a^2 + 2ab + b^2.$$
Subtract $2ab$ from both sides:
$$(a + b)^2 - 2ab = a^2 + b^2.$$
Done.
Proof 2 — Add to $(a-b)^2$.
Start with the other binomial-square identity:
$$(a - b)^2 = a^2 - 2ab + b^2.$$
Add $2ab$ to both sides:
$$(a - b)^2 + 2ab = a^2 + b^2.$$
The two proofs give the two equivalent forms.
Three Worked Examples, From Quick to Stretch
Quick. Find $3^2 + 4^2$.
$$3^2 + 4^2 = 9 + 16 = 25.$$
Cross-check using the formula: $(3 + 4)^2 - 2 \cdot 3 \cdot 4 = 49 - 24 = 25$. ✓
Final answer: $25$. (And, not coincidentally, $5^2 = 25$ — this is the famous $3$-$4$-$5$ right triangle.)
Standard (Wrong-Path-First). If $a + b = 7$ and $ab = 12$, find $a^2 + b^2$.
Wrong path. A student in our McKinney TX Grade 8 cohort once wrote: "$a + b = 7$ so $a = 3$ and $b = 4$; then $a^2 + b^2 = 9 + 16 = 25$." That works by luck — the values happen to fit — but the method generalises badly. If $a + b = 7$ and $ab = 10$, $a$ and $b$ are no longer integers; guessing fails.
Correct. Apply the formula directly:
$$a^2 + b^2 = (a + b)^2 - 2ab = 7^2 - 2 \cdot 12 = 49 - 24 = 25.$$
Final answer: $25$.
The method works for any $a + b$ and $ab$, integer or not. That's the value of using the formula instead of solving for $a$ and $b$ individually.
Stretch. A right triangle has legs in the ratio $5:12$ and hypotenuse $26$. Find the legs.
Let the legs be $5k$ and $12k$. By Pythagoras:
$$(5k)^2 + (12k)^2 = 26^2.$$ $$25k^2 + 144k^2 = 676.$$ $$169k^2 = 676.$$ $$k^2 = 4 \implies k = 2.$$
So the legs are $5k = 10$ and $12k = 24$.
Final answer: Legs are $10$ and $24$.
Cross-check: $10^2 + 24^2 = 100 + 576 = 676 = 26^2$. ✓ The triangle $(10, 24, 26)$ is a Pythagorean triple — $2$ times the primitive $(5, 12, 13)$.
Where the Formula Lives — From Right Triangles to Complex Numbers
The sum-of-two-squares identity is the algebraic shadow of distance in the plane.
Pythagorean theorem. For a right triangle with legs $a, b$ and hypotenuse $c$: $a^2 + b^2 = c^2$. Every distance computation in 2D Cartesian geometry collapses to this identity.
Complex numbers. The squared modulus of $a + bi$ is $|a + bi|^2 = a^2 + b^2$. Over the complex numbers the identity does factor: $a^2 + b^2 = (a + bi)(a - bi)$. This is the gateway to factoring quadratics whose discriminant is negative.
Vector magnitude in 2D. A vector $\vec{v} = (a, b)$ has length $|\vec{v}| = \sqrt{a^2 + b^2}$. The formula tells you how to evaluate without knowing $a, b$ individually — only their sum and product.
Pythagorean triples. Triples like $(3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25)$ satisfy $a^2 + b^2 = c^2$ with all three integers. Number theorists generated these systematically — there are infinitely many primitive ones.
Sum-of-two-squares theorem (Fermat). A prime $p$ can be written as $a^2 + b^2$ if and only if $p = 2$ or $p \equiv 1 \pmod{4}$. Primes like $5 = 1^2 + 2^2$, $13 = 2^2 + 3^2$, $17 = 1^2 + 4^2$ qualify; $7, 11, 19$ do not.
The identity is small. Its mathematical descendants span four centuries of number theory.
Three Habits That Lose Marks on a² + b²
1. Confusing $a^2 + b^2$ with $(a + b)^2$.
Where it slips in: Problems where the parentheses look similar.
Don't do this: Write $a^2 + b^2 = (a + b)^2$.
The correct way: $(a + b)^2 = a^2 + 2ab + b^2$ — bigger than $a^2 + b^2$ by the cross-term $2ab$. The sum-of-squares does not equal the square-of-sum unless $ab = 0$.
2. Trying to factor $a^2 + b^2$ over the real numbers.
Where it slips in: Polynomial factorization problems on test paper.
Don't do this: Write $a^2 + b^2 = (a + b)(a - b)$ — which is the difference-of-squares formula, not the sum.
The correct way: $a^2 + b^2$ does not factor over the real numbers. Its only real form is the identity $(a+b)^2 - 2ab$. Over complex numbers, $a^2 + b^2 = (a + bi)(a - bi)$. Don't force a real factorization that doesn't exist. Roughly five out of every ten Grade 8 students in our Bhanzu cohorts try to factor $a^2+b^2$ as $(a+b)(a-b)$ on the first attempt — the fix is to memorise sum is irreducible over the reals, difference factors.
3. Forgetting the $2$ in $2ab$.
Where it slips in: Substitution problems given $a + b$ and $ab$.
Don't do this: Write $a^2 + b^2 = (a + b)^2 - ab$.
The correct way: $a^2 + b^2 = (a + b)^2 - 2ab$. The factor of $2$ comes from the binomial-square expansion's cross-term $2ab$. Forgetting it is the same error class as forgetting the $2$ in $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca)$.
4. Using the formula on $a^2 - b^2$ or $a^3 + b^3$.
Where it slips in: Problems that look similar but involve different operations or powers.
Don't do this: Apply the sum-of-squares formula to factor $a^2 - b^2$ or $a^3 + b^3$.
The correct way: $a^2 - b^2 = (a-b)(a+b)$ — difference of squares, does factor. $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ — sum of cubes, factors. $a^2 + b^2$ — sum of squares, does not factor over the reals. Each identity is its own object. A real-world version of this error class: the 1996 Ariane 5 rocket failed because reused software applied a formula valid for one rocket (Ariane 4) to a different one with different parameters. The math worked — but for a system the formula was no longer modeling.
The Mathematician Behind a² + b²
The Pythagorean theorem — the geometric face of the $a^2 + b^2 = c^2$ identity — is named for Pythagoras of Samos (c. 570–c. 495 BCE, Greece), the philosopher-mathematician whose school taught geometry as a path to spiritual purification.
The historical record is messier than the school version. Babylonian clay tablets (notably Plimpton 322, c. 1800 BCE) list Pythagorean triples a millennium before Pythagoras was born — the result was known to surveyors and astronomers across Mesopotamia, India, and Egypt. The Indian mathematician Baudhāyana (c. 800 BCE) stated the theorem in Baudhāyana Śulbasūtra roughly $300$ years before Pythagoras. What Pythagoras's school may have contributed is the first systematic proof — though even that is uncertain, since the school's texts were oral and were destroyed.
The sum-of-two-squares identity in algebraic form — $a^2 + b^2 = (a+b)^2 - 2ab$ — is too elementary to attribute to a single discoverer. It was implicit in Euclid's Elements (Book II, Propositions 4 and 7, c. 300 BCE) as a geometric identity about square areas, and was used as a symbolic identity by every algebraist from al-Khwarizmi onward.
Conclusion
The a square plus b square formula is $a^2 + b^2 = (a + b)^2 - 2ab$, equivalently $(a - b)^2 + 2ab$.
It does not factor over the real numbers — unlike its cousin $a^2 - b^2 = (a-b)(a+b)$.
It is the algebraic form of the Pythagorean theorem when $a, b$ are the legs of a right triangle.
The most common mistake is treating $a^2 + b^2$ as equal to $(a+b)^2$ — it's smaller by $2ab$.
Over the complex numbers, $a^2 + b^2 = (a + bi)(a - bi)$ — the gateway to factoring negative-discriminant quadratics.
Practice These Three Before Moving On
Try these before continuing. If you slip on $(a+b)^2$ vs $a^2+b^2$, come back to Mistake 1.
Given $a + b = 10$ and $ab = 21$, find $a^2 + b^2$.
A right triangle has legs $9$ and $40$. Find the hypotenuse.
Verify that $5 = 1^2 + 2^2$ and find a similar two-square representation of $13$.
Want a live Bhanzu trainer to walk through more a square plus b square formula problems with your child? Book a free demo class — online globally.
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