You learn the identity x² - y² = (x - y)(x + y) early, and it feels like a magic key. Your brain wants to generalise: surely if a difference of squares splits, a sum of squares splits too? You try (x + y)(x + y), you try (x - y)(x + y), you guess. None give you x² + y².

That is not your brain failing. That is the algebra refusing. Over the real numbers, x² + y² does not factor. This is a hard structural fact about real-coefficient polynomials, not a "we haven't taught it yet". The earlier you accept it, the fewer marks you lose.

Why x² + y² has no real factorisation

Suppose x² + y² did split as a product (ax + by)(cx + dy) with real coefficients. Expand:

(ax + by)(cx + dy) = acx² + (ad + bc)xy + bdy².

For this to equal x² + 0·xy + y², three equations must hold: ac = 1, bd = 1, and ad + bc = 0.

From ac = 1 and bd = 1, both pairs (a, c) and (b, d) have positive product, so each pair is same-sign. Push the cases through ad + bc = 0 and you find ad and bc always end up with the same sign — their sum can never be zero.

Contradiction. No real a, b, c, d solve this system. The factorisation simply does not exist over the reals.

In complex numbers the story changes. With i = √(-1), the identity x² + y² = (x + iy)(x - iy) works beautifully. But the moment you write i, you have left the real numbers behind.

Test by attempting all simple factorisations

If you do not believe the algebra, try every plausible candidate by hand:

  • (x + y)(x + y) = x² + 2xy + y². Stray 2xy. Not x² + y².
  • (x - y)(x - y) = x² - 2xy + y². Cross term still there.
  • (x + y)(x - y) = x² - y². The famous one — gives the difference, not the sum.
  • (x + ky)(x + (1/k)y) = x² + (k + 1/k)xy + y² for non-zero real k. The middle coefficient k + 1/k is at least 2 in absolute value, never 0.

Every real candidate produces an unwanted middle term. The cross term is the obstruction.

To sharpen your eye, place x² + y² next to its cousins:

  • Difference of squares: x² - y² = (x - y)(x + y). Factors.
  • Difference of cubes: x³ - y³ = (x - y)(x² + xy + y²). Factors. The trinomial x² + xy + y² is itself irreducible over the reals.
  • Sum of cubes: x³ + y³ = (x + y)(x² - xy + y²). Factors.
  • Sum of squares: x² + y². Does NOT factor over the reals.
  • Sum of fourth powers: x⁴ + y⁴. Does NOT factor into clean linear or rational-coefficient quadratic pieces (a surd factorisation (x² + xy√2 + y²)(x² - xy√2 + y²) exists, but the standard school answer is "irreducible").

The pattern — odd vs even

Watch the symmetry:

  • Differences x^n - y^n always factor for n ≥ 2, regardless of whether n is odd or even. (x - y) is always a factor.
  • Sums x^n + y^n factor over the reals only when n is odd. For even n > 1, no factorisation with linear real-coefficient pieces exists.

So x³ + y³, x⁵ + y⁵, x⁷ + y⁷ all split. But x² + y², x⁴ + y⁴, x⁶ + y⁶ resist.

Why the odd sum works, even sum doesn't

There is a clean reason rooted in roots. To check whether (x + y) divides x^n + y^n, set x = -y:

  • For odd n: (-y)^n + y^n = -y^n + y^n = 0. So x = -y is a root, meaning (x + y) divides x^n + y^n.
  • For even n: (-y)^n + y^n = 2y^n. Not zero. So (x + y) is not a factor, and in fact x^n + y^n has no real roots at all (with y fixed and non-zero).

No real roots means no linear real factors. That is the deep reason x² + y² resists.

Students' wrong attempts

Three classic errors show up on test papers every year:

  • x² + y² = (x + y)². Wrong — that gives x² + 2xy + y².
  • x² + y² = (x - y)(x + y). Wrong — that gives x² - y².
  • x² + y² = (xi + y)(-xi + y). Algebraically valid, but uses i. Not a real factorisation.

At the school level, the expected answer is "irreducible over the real numbers". Writing (x + iy)(x - iy) only earns marks at the complex-numbers stage.

What if you're asked to "factor x² + y²"

Read the question carefully. If the chapter is real-coefficient factorisation, the answer is "cannot be factored". If the chapter is complex numbers, write (x + iy)(x - iy). The setting decides.

Sums of squares show up in many disguises. Each of these is irreducible over the reals:

  • x² + 1: x² + y² with y = 1.
  • x² + 4: x² + (2)².
  • x² + x + 1: discriminant 1 - 4 = -3 < 0. No real roots, no real linear factors.
  • x² + y²: the archetype.

Whenever you see a quadratic with negative discriminant or a pure sum of even powers, this thing does not split over the reals.

Complex-number factorisation (for context)

Complex numbers cure everything. With i² = -1, x² + y² = (x + iy)(x - iy). Verify: (x + iy)(x - iy) = x² - (iy)² = x² - i²y² = x² + y². Tick.

This is part of the Fundamental Theorem of Algebra: every positive-degree polynomial splits into linear factors over the complex numbers. Over the reals you only get linear and irreducible-quadratic pieces — and x² + y² is the canonical irreducible quadratic.

Trick problems involving x² + y²

A classic Indian textbook problem: "If x + y = 5 and xy = 6, find x² + y²."

The question contains x² + y² but is not asking you to factor it — only to evaluate it. Use (x + y)² = x² + 2xy + y²: 25 = x² + y² + 12, giving x² + y² = 13. Done, without pretending it factors. The trick is recognising it as part of a binomial-square identity.

Recognition drill

Quick sort. For each, say "factors over reals" or "does not":

  • x² - y² → factors as (x - y)(x + y).
  • x² + y² → does not factor over reals.
  • x³ + y³ → factors as (x + y)(x² - xy + y²).
  • x³ - y³ → factors as (x - y)(x² + xy + y²).
  • x⁴ - y⁴ → (x² - y²)(x² + y²) = (x - y)(x + y)(x² + y²). First two pieces split; the last refuses.
  • x² + 4y² → does not factor (sum of squares in disguise: x² + (2y)²).
  • x² + 2xy + y² - 9 → (x + y - 3)(x + y + 3). Spot (x + y)² - 9, then difference of squares.

The last one reminds you: the question is not always "does this raw shape factor?" It is "can I rewrite it in a form that factors?" Clever grouping rescues many expressions. But raw x² + y² is unsalvageable.

Closing

Differences of squares factor. Sums of squares do not — over the real numbers. That asymmetry is one of the cleanest, most useful facts in school algebra, and it generalises: differences of any power factor; sums factor only for odd powers.

Never write (x + y)(x - y) when you mean x² + y². They are not the same. They differ by a sign in the middle, and that sign decides whether the expression splits or stays stubbornly whole.