Prove polynomial identities and use them to describe numerical relationships such as Pythagorean triples

Learning goal: Prove polynomial identities by transforming one side into the other, distinguish an identity from an equation that merely has solutions, and use identities to describe numerical relationships, including generating Pythagorean triples from the identity (x² + y²)² = (x² − y²)² + (2xy)².

What an identity is. An identity is an equation that is true for every value of its variables. The equation (a + b)² = a² + 2ab + b² is an identity: whatever numbers a and b are, the two sides are equal. Compare it with x² = 9, which is true only for x = 3 and x = −3; that is an equation to be solved, not an identity. The distinction matters because an identity can be used as a rule: wherever the left side appears, it may be replaced by the right side. The identities from earlier work are (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and a² − b² = (a − b)(a + b). Two more are used in this lesson: (a + b)³ = a³ + 3a²b + 3ab² + b³ and a³ − b³ = (a − b)(a² + ab + b²).

How to prove an identity. A proof shows that the two sides are the same polynomial. The usual method is to expand one side, or both, and collect like terms until the expressions match term for term. A proof is not a set of numerical checks. Substituting a = 1 and b = 2 into (a + b)² = a² + b² gives 9 = 5, which shows that equation is false, so checks are excellent at disproving; but substituting a = 0 gives b² = b², true for every b, and no finite list of such checks can by itself show an equation holds for all values. (Two polynomials in one variable, each of degree at most d, that agree at d + 1 different inputs are in fact identical, which is a theorem of later mathematics; it does not carry over simply to two variables.) The standard's intended method, and the one used here, is algebraic expansion.

A fully worked model: the Pythagorean-triple identity. Prove that (x² + y²)² = (x² − y²)² + (2xy)² for all x and y. Left side: (x² + y²)² is a square of a sum, so it equals (x²)² + 2(x²)(y²) + (y²)² = x⁴ + 2x²y² + y⁴. Right side, first term: (x² − y²)² = x⁴ − 2x²y² + y⁴. Second term: (2xy)² = 4x²y². Add them: x⁴ − 2x²y² + y⁴ + 4x²y² = x⁴ + 2x²y² + y⁴. The two sides are the same polynomial, so the identity is proved.

Using the identity to describe a numerical relationship. A Pythagorean triple is a set of three positive whole numbers a, b, c with a² + b² = c². The identity says that for any x and y, the numbers x² − y², 2xy and x² + y² satisfy (x² − y²)² + (2xy)² = (x² + y²)², so they form a Pythagorean triple whenever they are positive whole numbers, which happens for whole numbers x > y > 0. Take x = 2, y = 1: x² − y² = 3, 2xy = 4, x² + y² = 5, and indeed 9 + 16 = 25. Take x = 3, y = 2: 5, 12, 13, and 25 + 144 = 169. Take x = 4, y = 1: 15, 8, 17, and 225 + 64 = 289. Take x = 4, y = 3: 7, 24, 25, and 49 + 576 = 625. The table in the diagram lists these. Some choices give multiples of smaller triples: x = 3, y = 1 gives 8, 6, 10, which is twice 4, 3, 5. The identity does not promise a new triple each time, but it promises a triple every time, and that is what "use them to describe numerical relationships" means: a single algebraic identity explains an infinite family of numerical facts.

A second identity and its numerical use. The identity (n + 1)² − n² = 2n + 1 is proved by expanding: n² + 2n + 1 − n² = 2n + 1. It says the difference between consecutive squares is always the odd number 2n + 1. So 25² − 24² = 2·24 + 1 = 49, without computing either square. Likewise the difference-of-squares identity explains mental arithmetic: 41 × 39 = (40 + 1)(40 − 1) = 40² − 1 = 1599, and 103² = (100 + 3)² = 10000 + 600 + 9 = 10609.

Proving the cube identity. Show that (a − b)(a² + ab + b²) = a³ − b³. Distribute a: a³ + a²b + ab². Distribute −b: −a²b − ab² − b³. Add: a³ + (a²b − a²b) + (ab² − ab²) − b³ = a³ − b³. The middle terms cancel in pairs, and the identity is proved. It explains why 1000 − 27 = 973 must be divisible by 10 − 3 = 7: 973 = 7 × 139.

The misconception to remove. Students sometimes "prove" an identity by starting from the equation, manipulating both sides at once and arriving at 0 = 0. That argument runs backward and can prove false statements if a step is not reversible (squaring both sides, for instance). The safe form of proof is one-directional: begin with one side, transform it by expansions and collecting, and end at the other side. A second misconception is that (a + b)² = a² + b²; the single check a = b = 1 gives 4 = 2 and settles it.

Why identities matter beyond this lesson. Completing the square, factoring, the quadratic formula and the Binomial Theorem are all identities or consequences of identities. Every time you rewrite an expression in an equivalent form, you are using one. Knowing how to prove them means knowing when a rewrite is legitimate.

Worked proof of the cube of a sum: Prove (a + b)³ = a³ + 3a²b + 3ab² + b³. Write (a + b)³ = (a + b)(a + b)². Using the square identity, (a + b)² = a² + 2ab + b². Now distribute: a(a² + 2ab + b²) = a³ + 2a²b + ab², and b(a² + 2ab + b²) = a²b + 2ab² + b³. Add: a³ + (2a²b + a²b) + (ab² + 2ab²) + b³ = a³ + 3a²b + 3ab² + b³. Both sides are now the same polynomial. Numerical use: 11³ = (10 + 1)³ = 1000 + 300 + 30 + 1 = 1331.

Worked triple from the identity: Generate the Pythagorean triple for x = 5, y = 2. Compute x² − y² = 25 − 4 = 21, 2xy = 2·5·2 = 20 and x² + y² = 25 + 4 = 29. Check: 21² + 20² = 441 + 400 = 841 and 29² = 841. The triple is 20, 21, 29. The check is a confirmation of one instance; the guarantee that it would work came from the proved identity.

Worked identity for a sum of squares: Prove (a + b)² + (a − b)² = 2a² + 2b². Expand: (a² + 2ab + b²) + (a² − 2ab + b²). The 2ab terms cancel and the rest collects to 2a² + 2b². Numerical relationship: for any two numbers, the squares of their sum and their difference add to twice the sum of their squares. For 7 and 3: 10² + 4² = 100 + 16 = 116, and 2·49 + 2·9 = 98 + 18 = 116.

Worked disproof by substitution: Is (x + 1)² = x² + 1 an identity? At x = 1 the left side is 4 and the right side is 2. One failing value is enough to show it is not an identity. It is, however, an equation with a solution: expanding gives x² + 2x + 1 = x² + 1, so 2x = 0 and x = 0 is the only value where it holds.

Representation task: Make a table with the columns x, y, x² − y², 2xy, x² + y², and "check a² + b² = c²". Fill it for the pairs (2, 1), (3, 2), (4, 1), (4, 3), (5, 2) and one pair of your own with whole numbers x > y > 0. In the last column write the two squares and their sum against the third square. Beneath the table, write out the full algebraic proof of (x² + y²)² = (x² − y²)² + (2xy)², one expansion per line, and circle the line where the two sides are seen to be the same polynomial.

Independent performance: Prove two identities in writing, starting from one side and ending at the other: a² − b² = (a − b)(a + b) and a³ + b³ = (a + b)(a² − ab + b²). For each, give one numerical relationship it describes, with a specific computation (for example, a product of two numbers found without a calculator, or a divisibility fact). Then write a short paragraph explaining why a table of successful checks does not prove an identity, and why one failed check does disprove one. Success criteria: each proof is one-directional and every expansion step is shown; the numerical examples are correctly computed; the paragraph names the difference between "true for some values" and "true for all values".