---
title: "The Pythagorean Theorem"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 58
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem
---

# Chapter 58 — The Pythagorean Theorem

The most famous theorem of all mathematics relates the three sides of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles). With it, lengths become computable that no ruler could measure directly: diagonals, distances, heights. It is also our first big encounter with the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) between a theorem and its converse.

## 58.1 The theorem

**Theorem 58.1 (Pythagoras).**

If a triangle $ABC$ is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$, then

$$
BC^2 = AB^2 + AC^2 :
$$

the square of the hypotenuse (the side opposite the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle)) equals the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the squares of the other two sides.

**Idea of proof.** Four copies of the triangle, arranged inside a big square of side $AB + AC$, leave uncovered a tilted square built on the hypotenuse: comparing [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) gives the equality. The detailed computation is proposed as [Exercise 58.11](#exo-g8-pythagoras-11); the theorem also drops out of the scalar [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def), studied in the High School volume. ∎

![The theorem as an area statement: the square built on the hypotenuse (25) has exactly the same area as the two other squares together (16 + 9). Here the sides are 3, 4, 5.](https://one-course.com/images/onecourse/chapters/math-1/g8-pythagoras/fig-1633da61d8e5.svg)

*The theorem as an [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) statement: the square built on the hypotenuse ($25$) has exactly the same [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) as the two other squares together ($16 + 9$). Here the sides are $3$, $4$, $5$.*

**Method 58.2 (Computing a length).**

In a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) where two sides are known:

1. name the hypotenuse (opposite the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) — always the longest side);
2. write the equality of [Theorem 58.1](#thm-g8-pythagoras-direct) with the known values;
3. solve for the missing square: *add* the squares if the hypotenuse is unknown, *subtract* if a leg is unknown;
4. take the square root, exactly or with a calculator, and check that the hypotenuse came out longest.

**Example 58.3 (Finding the hypotenuse).**

A [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has legs $6$ and $8$. Hypotenuse $c$:

$$
c^2 = 6^2 + 8^2 = 36 + 64 = 100,
\qquad
c = \sqrt{100} = 10 .
$$

(The number $\sqrt{100}$, the *square root* of $100$, is the positive number whose square is $100$; square roots get a full chapter in [Chapter 65](https://one-course.com/books/math/1/en/chapter/65-square-roots#ch-g9-sqrt).)

**Example 58.4 (Finding a leg).**

A $5$ m ladder leans against a wall, its foot $1.4$ m from the wall. Height reached $h$: the ladder is the hypotenuse, so

$$
h^2 = 5^2 - 1.4^2 = 25 - 1.96 = 23.04,
\qquad
h = \sqrt{23.04} = 4.8 \text{ m}.
$$

Subtract, not add: the unknown here is a *leg*.

## 58.2 The converse

**Theorem 58.5 (Converse of Pythagoras).**

If, in a triangle $ABC$, the sides satisfy $BC^2 = AB^2 + AC^2$, then the triangle is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$.

**Proof.** *Admitted at this level.* ∎

**Method 58.6 (Testing a right angle).**

Given the three sides of a triangle:

1. compute separately the square of the longest side, and the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the squares of the two others;
2. if the two results are equal, the triangle is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) (converse), at the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) opposite the longest side;
3. if they differ, it is not [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) — because Pythagoras’ theorem would force equality.

**Example 58.7.**

Sides $5$, $12$, $13$: longest side squared, $13^2 = 169$; [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the other squares, $5^2 + 12^2 = 25 + 144 = 169$. Equal: [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) (opposite the side $13$).

Sides $6$, $7$, $9$: $9^2 = 81$ but $6^2 + 7^2 = 85 \neq 81$: not a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) (it is “almost” one — the numbers decide, not the eye).

**Remark 58.8 (Theorem vs converse).**

The theorem and its converse say different things: the theorem *uses* a [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) to compute lengths; the converse uses lengths to *prove* a [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle). Builders have used the converse for millennia: a rope with knots at $3$, $4$, $5$ units pulled taut gives a perfect [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle).

**Example 58.9 (Diagonal of a square).**

The diagonal of a square of side $1$ is the hypotenuse of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs $1$ and $1$:

$$
d^2 = 1^2 + 1^2 = 2 ,
$$

so $d = \sqrt2 \approx 1.414$ — a number that is not a [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def), as the High School volume will prove. For a square of side $c$, the diagonal is $c\sqrt2$.

## 58.3 Exercises

**Exercise 58.1 ★.**

In each [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), compute the hypotenuse: legs $9$ and $12$; legs $5$ and $12$; legs $8$ and $15$.

**Solution of Exercise 58.1.**

$\sqrt{81 + 144} = \sqrt{225} = 15$; $\sqrt{25 + 144} = \sqrt{169} = 13$; $\sqrt{64 + 225} = \sqrt{289} = 17$.

**Exercise 58.2 ★.**

A [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has hypotenuse $25$ and one leg $7$. Compute the other leg.

**Solution of Exercise 58.2.**

Leg$^2 = 25^2 - 7^2 = 625 - 49 = 576$, so the leg is $\sqrt{576} = 24$.

**Exercise 58.3 ★.**

Compute the diagonal of a rectangle of sides $12$ cm and $9$ cm; the diagonal of a square of side $5$ cm (exact value with a square root, then rounded to the mm).

**Solution of Exercise 58.3.**

Rectangle: $\sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$ cm.

Square: $d = 5\sqrt2 \approx 7.1$ cm ([Example 58.9](#ex-g8-pythagoras-diagonal)).

**Exercise 58.4 ★.**

Which triangles are [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)? Justify with [Method 58.6](#met-g8-pythagoras-test):

$$
(20,\ 21,\ 29); \qquad
(7,\ 8,\ 11); \qquad
(1.5,\ 2,\ 2.5).
$$

**Solution of Exercise 58.4.**

$(20, 21, 29)$: $29^2 = 841$ and $20^2 + 21^2 = 400 + 441 = 841$: [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

$(7, 8, 11)$: $121 \neq 49 + 64 = 113$: not [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles).

$(1.5, 2, 2.5)$: $6.25 = 2.25 + 4$: [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) (it is $(3,4,5)$ halved).

**Exercise 58.5 ★.**

A gate is braced by a diagonal plank. The gate is $3$ m wide and $1.6$ m tall: how long is the plank?

**Solution of Exercise 58.5.**

Plank $= \sqrt{3^2 + 1.6^2} = \sqrt{9 + 2.56} = \sqrt{11.56} = 3.4$ m.

**Exercise 58.6 ★.**

An [isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) has two sides of $10$ cm and a base of $12$ cm. Its height splits it into two [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles): compute that height, then the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the triangle.

**Solution of Exercise 58.6.**

The height falls on the midpoint of the base ([isosceles triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles)), so each half-triangle has hypotenuse $10$ and one leg $6$: $h = \sqrt{100 - 36} = \sqrt{64} = 8$ cm. [Area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area): $\frac{12 \times 8}{2} = 48$ cm$^2$.

**Exercise 58.7 ★★.**

A television screen is a $16{:}9$ rectangle measuring $88.5$ cm by $49.8$ cm. Screens are sold by the diagonal, in inches ($1$ inch $= 2.54$ cm). Compute the diagonal in cm, then in inches ([round](https://one-course.com/books/math/1/en/chapter/21-large-numbers#def-g4-numbers-round) to the nearest inch).

**Solution of Exercise 58.7.**

Diagonal: $\sqrt{88.5^2 + 49.8^2} = \sqrt{7832.25 + 2480.04}
= \sqrt{10312.29} \approx 101.5$ cm. In inches: $101.5 \div 2.54 \approx 40$ inches.

**Exercise 58.8 ★★.**

A boat sails $24$ km east, then $10$ km north. How far is it from its starting point (as the crow flies)? Draw the situation first.

**Solution of Exercise 58.8.**

The two legs are $24$ km and $10$ km: $\sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26$ km.

**Exercise 58.9 ★★.**

A $2.5$ m ladder must reach a window sill $2.4$ m above the ground. How far from the wall must its foot be placed? Is the answer compatible with safety advice (foot at about a [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the ladder length from the wall)?

**Solution of Exercise 58.9.**

Foot distance: $\sqrt{2.5^2 - 2.4^2} = \sqrt{6.25 - 5.76} =
\sqrt{0.49} = 0.7$ m. A [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the ladder is $2.5 \div 4 \approx 0.6$ m: the $0.7$ m found is close to the advised placement — acceptable.

**Exercise 58.10 ★★.**

On a coordinate grid, plot $A(1, 2)$ and $B(5, 5)$, and draw the [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the axes whose hypotenuse is $[AB]$. Compute $AB$. (This construction, for any two points, becomes the distance formula of coordinate geometry, in the High School volume.)

**Solution of Exercise 58.10.**

The legs measure $5 - 1 = 4$ (horizontal) and $5 - 2 = 3$ (vertical), so

$$
AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 .
$$

**Exercise 58.11 ★★★.**

Four copies of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs $a$, $b$ and hypotenuse $c$ are placed in the corners of a square of side $a + b$, leaving a tilted square of side $c$ uncovered in the middle.

1. Express the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the big square in two ways: directly, and as (four triangles) $+$ (tilted square).
2. Expand $(a + b)^2$ (see [Theorem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#thm-g8-equations-expand) ) and deduce $c^2 = a^2 + b^2$ : you have proved the Pythagorean theorem.

**Solution of Exercise 58.11.**

*1.* Directly: $(a + b)^2$. As pieces: four triangles of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $\frac{ab}{2}$ each, plus the tilted square $c^2$:

$$
(a + b)^2 = 4 \times \frac{ab}{2} + c^2 = 2ab + c^2 .
$$

*2.* Expanding the left side: $(a+b)^2 = a^2 + 2ab + b^2$. So

$$
a^2 + 2ab + b^2 = 2ab + c^2
\quad\Longrightarrow\quad
a^2 + b^2 = c^2 . \qed
$$

## 58.4 Problem: Pythagorean triples

**Problem 58.1.**

Weekend problem — whole-number right triangles, and the identity $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ that generates them all

A *Pythagorean triple* is a triple of whole numbers $(a, b, c)$ with

$$
a^2 + b^2 = c^2 ,
$$

like the builders’ $(3, 4, 5)$. A Babylonian clay tablet catalogued huge ones — $(4601,\ 4800,\ 6649)$ among them — more than a thousand years before Pythagoras. This problem hunts for triples, then builds the machine that produces them: an algebraic identity straight out of [Chapter 57](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#ch-g8-equations), put in the service of geometry.

**Part I — Hunting for triples.**

1. Check that $(5, 12, 13)$ , $(8, 15, 17)$ and $(20, 21, 29)$ are [Pythagorean triples](#pb-g8-pythagoras-1) . What does [Theorem 58.5](#thm-g8-pythagoras-converse) say about triangles with these side lengths?
2. Explain why a taut rope with twelve equal [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) , laid out as a triangle with sides of $3$ , $4$ and $5$ [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) , gives builders a perfect [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) .
3. Show that if $(a, b, c)$ is a [Pythagorean triple](#pb-g8-pythagoras-1) , so is $(ka, kb, kc)$ for every whole number $k \geq 1$ . Deduce three new triples from $(3, 4, 5)$ .
4. The [multiples](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $(3, 4, 5)$ are the triples $(3k, 4k, 5k)$ . Show that $(5, 12, 13)$ is *not* one of them.
5. Explain why questions 3 and 4 together prove that scaling $(3, 4, 5)$ can never produce every [Pythagorean triple](#pb-g8-pythagoras-1) : some other machine is needed.

**Part II — The machine.** Take two whole numbers $m > n \geq 1$ and form the three numbers

$$
a = m^2 - n^2,
\qquad
b = 2mn,
\qquad
c = m^2 + n^2 .
$$

6. Using the remarkable identities of [Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1) (write $A = m^2$ and $B = n^2$), expand $a^2$ and $b^2$, and prove that $$\left(m^2 - n^2\right)^2 + (2mn)^2  = \left(m^2 + n^2\right)^2 :$$ the machine always outputs a [Pythagorean triple](#pb-g8-pythagoras-1).
7. Run the machine on $(m, n) = (2, 1)$ , $(3, 1)$ , $(3, 2)$ , $(4, 1)$ and $(4, 3)$ , and display the five triples in a table.
8. One triple in your table is a scaled copy of a smaller one. Which, and of which?
9. Find $(m, n)$ for which the machine outputs the triple $(20, 21, 29)$ of question 1 (legs in either order).
10. Find $m$ and $n$ with $m^2 + n^2 = 100$ , and deduce a [Pythagorean triple](#pb-g8-pythagoras-1) whose hypotenuse is $100$ .

**Part III — Families, and a parity finale.**

11. Run the machine with $m = n + 1$ . Show that the [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg is $2n + 1$ , the [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg $2n^2 + 2n$ , and the hypotenuse $2n^2 + 2n + 1$ — so hypotenuse and [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg differ by exactly $1$ . Write out the triples for $n = 1, 2, 3, 4$ .
12. Deduce that *every* [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $3, 5, 7, 9, \dots$ is the side of some whole-number [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) , and give a triple containing $11$ .
13. [Even numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) work too: run the machine with $n = 1$ , and produce a triple containing $14$ . Then obtain the same triple a second time, by scaling one from your table of question 7.
14. Show that the square of an [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) and that the square of an [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) . (Write the number as $2k$ or $2k + 1$ and use an identity.)
15. Conclude the parity finale: there is no [Pythagorean triple](#pb-g8-pythagoras-1) with all three numbers [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) — in every whole-number [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) , at least one side is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) .

**Solution of Problem 58.1.**

**1.** $5^2 + 12^2 = 25 + 144 = 169 = 13^2$; $8^2 + 15^2 = 64 + 225 = 289 = 17^2$; $20^2 + 21^2 = 400 + 441 = 841 = 29^2$. All three are triples, and by [Theorem 58.5](#thm-g8-pythagoras-converse) a triangle with these side lengths is [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles), the [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) facing the longest side.

**2.** The rope forms a triangle with sides $3$, $4$, $5$ (in [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects)), and $3^2 + 4^2 = 25 = 5^2$: by the converse ([Theorem 58.5](#thm-g8-pythagoras-converse)), the angle between the sides of $3$ and $4$ [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) is exactly right — no protractor needed, only knots.

**3.** If $a^2 + b^2 = c^2$, then

$$
(ka)^2 + (kb)^2 = k^2 a^2 + k^2 b^2
= k^2\left(a^2 + b^2\right) = k^2 c^2 = (kc)^2 .
$$

From $(3, 4, 5)$: for instance $(6, 8, 10)$, $(9, 12, 15)$ and $(30, 40, 50)$.

**4.** A [multiple](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) $(3k, 4k, 5k)$ containing $5$ as its smallest number would need $3k = 5$, and no whole number $k$ satisfies that ($k = 1$ gives $3$, $k = 2$ gives $6$). So $(5, 12, 13)$ is not a [multiple](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $(3, 4, 5)$.

**5.** Question 3 shows scaling produces only the triples $(3k, 4k, 5k)$; question 4 exhibits a genuine triple that is not of this form. So the family of all [Pythagorean triples](#pb-g8-pythagoras-1) is strictly larger than the [multiples](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $(3, 4, 5)$: producing them requires something new.

**6.** Write $A = m^2$ and $B = n^2$. By the remarkable identities ([Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1)):

$$
a^2 = (A - B)^2 = A^2 - 2AB + B^2,
\qquad
b^2 = (2mn)^2 = 4m^2n^2 = 4AB .
$$

Adding:

$$
a^2 + b^2 = A^2 - 2AB + B^2 + 4AB
= A^2 + 2AB + B^2 = (A + B)^2
= \left(m^2 + n^2\right)^2 = c^2 .
$$

**7.**

| $(m, n)$ | $m^2 - n^2$ | $2mn$ | $m^2 + n^2$ |
| --- | --- | --- | --- |
| $(2, 1)$ | $3$ | $4$ | $5$ |
| $(3, 1)$ | $8$ | $6$ | $10$ |
| $(3, 2)$ | $5$ | $12$ | $13$ |
| $(4, 1)$ | $15$ | $8$ | $17$ |
| $(4, 3)$ | $7$ | $24$ | $25$ |

**8.** $(8, 6, 10)$ is the double of $(4, 3, 5)$ — that is, of the builders’ $(3, 4, 5)$. The machine sometimes rediscovers an old triple in disguise.

**9.** We need $m^2 - n^2 = 21$ and $2mn = 20$, so $mn = 10$: trying $m = 5$, $n = 2$ gives $25 - 4 = 21$ and $2 \times 10 = 20$, with hypotenuse $25 + 4 = 29$. The machine outputs $(21, 20, 29)$ for $(m, n) = (5, 2)$.

**10.** $m = 8$, $n = 6$ gives $m^2 + n^2 = 64 + 36 = 100$. The triple is

$$
\left(64 - 36,\ 2 \times 48,\ 100\right) = (28,\ 96,\ 100),
$$

and indeed $28^2 + 96^2 = 784 + 9216 = 10\,000 = 100^2$.

**11.** With $m = n + 1$:

$$
m^2 - n^2 = (n + 1)^2 - n^2 = 2n + 1
$$

([difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of consecutive squares, [Problem 57.1](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#pb-g8-equations-1)); $2mn = 2n(n + 1) = 2n^2 + 2n$; and $m^2 + n^2 = n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1$. The hypotenuse exceeds the [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg by exactly $1$. For $n = 1, 2, 3, 4$:

$$
(3, 4, 5), \quad (5, 12, 13), \quad (7, 24, 25), \quad
(9, 40, 41) .
$$

**12.** Every [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) at least $3$ is $2n + 1$ for some $n \geq 1$, and question 11 provides a triple whose [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg is exactly $2n + 1$. For $11 = 2 \times 5 + 1$, take $n = 5$:

$$
(11,\ 60,\ 61),
\qquad
11^2 + 60^2 = 121 + 3600 = 3721 = 61^2 .
$$

**13.** With $n = 1$ the machine gives $\left(m^2 - 1,\ 2m,\ m^2 + 1\right)$, whose [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) leg $2m$ can be any [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) from $4$ upward. For $14 = 2 \times 7$, take $m = 7$: the triple $(48, 14, 50)$. The same triple appears by doubling $(7, 24, 25)$ from the table of question 7: $(14, 48, 50)$ — the same three numbers.

**14.** An [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is $2k$, and $(2k)^2 = 4k^2$ is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) (a [multiple](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $4$, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)). An [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is $2k + 1$, and the first remarkable identity gives

$$
(2k + 1)^2 = 4k^2 + 4k + 1 = 2\left(2k^2 + 2k\right) + 1 ,
$$

an [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) plus $1$: [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd).

**15.** Suppose all of $a$, $b$, $c$ were [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd). By question 14, $a^2$ and $b^2$ are [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), so $a^2 + b^2$ is a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of two [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd): *[even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)*. But $c^2$ would be *[odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)*. The equality $a^2 + b^2 = c^2$ is then impossible. So in every [Pythagorean triple](#pb-g8-pythagoras-1) at least one of the three numbers is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) — as every line of the table of question 7 confirms.