Core practice ← Back to lesson

Generating Pythagorean Triples: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 Three outputs from two inputs

Use the parameter pair $(13,10)$ to generate a Pythagorean triple.

Identify the hypotenuse and explain why the triple is primitive.
1. Hint 1
  The square sum gives the longest length.
2. Hint 2
  Check both the GCF and the parity of the input pair.
Answer

The triple is $(69,260,269)$ with hypotenuse $269$ . It is primitive because $13$ and $10$ are coprime and have opposite parity.

Full solution

Compute the difference of the squares to find one leg.
$13^2-10^2=69$

Double the product for the other leg.
$2\cdot13\cdot10=260$

Add the squares to find the hypotenuse.
$13^2+10^2=269$

The prime $13$ has no common factor greater than $1$ with $10$ , and one parameter is odd while the other is even.
The triple is primitive.

Answer

The triple is $(69,260,269)$ with hypotenuse $269$ . It is primitive because $13$ and $10$ are coprime and have opposite parity.

Key idea

The square sum is the hypotenuse, while both input tests determine primitivity.
Problem 2 The smallest integer copy

The right triangle shown has side lengths $352$ , $420$ , and $548$ units.

Find the primitive triple and the integer scale factor from the primitive triangle to the given one.

The drawing shows the lengths before you reduce the triple.

Text description of this figure

The right triangle has a horizontal leg labeled 420, a vertical leg labeled 352, and a hypotenuse labeled 548.
1. Hint 1
  Look for one factor that divides every label.
2. Hint 2
  After dividing, verify that the three quotients have no common factor greater than one.
Answer

The primitive triple is $(88,105,137)$ . The scale factor to the given triangle is $4$ .

Full solution

All three lengths are divisible by $4$ .
$\left(\frac{352}{4},\frac{420}{4},\frac{548}{4}\right)=(88,105,137)$

The prime factors of $88$ are $2$ and $11$ .
Neither divides $105$ , so no prime divides all three quotients.

The quotient has GCF $1$ , so the divisor $4$ was the full GCF of the original lengths and the scale factor is $4$ .

Answer

The primitive triple is $(88,105,137)$ . The scale factor to the given triangle is $4$ .

Key idea

Confirm that the quotient is primitive before calling a common factor the full scale factor.
Problem 3 Two ways to get a scaled output

For each case, state which primitive condition fails, find the GCF of the outputs, and give the primitive triple. The parameters are positive with $m>n$ .

Case A uses the parameter pair $(19,5)$ .

Case B uses the parameter pair $(24,21)$ .
1. Hint 1
  Check parity and common factors separately for each input pair.
2. Hint 2
  In Case B, remove the common factor from the inputs before using the three formulas.
3. Hint 3
  In Case A, the sum of two odd squares can be divided by two but not by four.
Answer

Case A fails opposite parity, has GCF $2$ , and reduces to $(168,95,193)$ . Case B fails coprimality, has GCF $9$ , and reduces to $(15,112,113)$ . Either order of the legs is valid.

Full solution

Case A.
The parameters are coprime but both odd, and the outputs are $(336,190,386)$ .
The square sum $386$ is even and has no factor of $4$ .

No odd prime can divide all Case A outputs because it would have to divide both parameter squares.
The GCF is $2$ , so division gives $(168,95,193)$ .

Case B.
The parameters have opposite parity but share GCF $3$ .
Dividing them by $3$ gives $(8,7)$ , which is coprime and of opposite parity.

The Case B outputs from $(8,7)$ are $(15,112,113)$ and have GCF $1$ .
Each entry generated by the original pair is nine times the corresponding entry, so the original GCF is $9$ .

Answer

Case A fails opposite parity, has GCF $2$ , and reduces to $(168,95,193)$ . Case B fails coprimality, has GCF $9$ , and reduces to $(15,112,113)$ . Either order of the legs is valid.

Key idea

A shared parameter factor enters the outputs squared, while same parity supplies a common factor of two.
Problem 4 Which leg goes into the half-difference?

The primitive right triangle shown has legs $105$ and $208$ units and hypotenuse $233$ units.

Recover the positive parameters $m>n$ . Verify the given even leg using your parameters.

The horizontal leg is shorter in this triangle.

Text description of this figure

The right triangle has a horizontal leg labeled 105, a vertical leg labeled 208, and a hypotenuse labeled 233.
1. Hint 1
  Use parity to select the leg that pairs with the hypotenuse.
2. Hint 2
  The half-sum and half-difference are the parameter squares. Take their positive roots before checking the product.
Answer

The parameters are $(13,8)$ . The doubled product is $208$ , the given even leg.

Full solution

Use the odd leg $105$ and the hypotenuse $233$ .
The sum gives the larger square.
$\frac{233+105}{2}=169$

The difference gives the smaller square.
$\frac{233-105}{2}=64$

Taking positive roots gives
$(m,n)=(13,8).$

The doubled product gives the even leg.
$2\cdot13\cdot8=208$

Answer

The parameters are $(13,8)$ . The doubled product is $208$ , the given even leg.

Key idea

The odd leg goes into the two half-operations regardless of which leg is longer.
Problem 5 Build the largest copy with a fixed stock

A model right-triangle frame has side lengths $84$ , $187$ , and $205$ cm. A shop will make a larger frame by multiplying each of the three model lengths by the same integer greater than one.

The shop has $1{,}500$ cm of stock, and cutting wastes no material. Find the greatest allowable scale factor, the three frame lengths, and the unused stock.

Every side of the model gets multiplied by the same integer.

Text description of this figure

The right triangle has a horizontal leg labeled 187 cm, a vertical leg labeled 84 cm, and a hypotenuse labeled 205 cm.
1. Hint 1
  The stock limit applies to the sum of all three lengths.
2. Hint 2
  Compare consecutive integer multiples of the model perimeter near the limit.
Answer

The greatest scale factor is $3$ . The frame lengths are $252$ , $561$ , and $615$ cm, and $72$ cm of stock remains.

Full solution

Add the three model lengths.
$84+187+205=476$

A scale factor of $3$ uses $1{,}428$ cm, and a scale factor of $4$ would use $1{,}904$ cm.
Larger factors only increase the perimeter.

Multiply each model length by $3$ to get $252$ , $561$ , and $615$ cm.

Subtract the stock used from the stock available.
$1500-1428=72$

Answer

The greatest scale factor is $3$ . The frame lengths are $252$ , $561$ , and $615$ cm, and $72$ cm of stock remains.

Key idea

The perimeter of an integer-scale copy grows by the same factor as its sides.
Problem 6 An input from a short list

For Euclid's formulas, fix $m=15$ . The other parameter $n$ must be one of $4$ , $6$ , $7$ , or $10$ .

Select every choice that makes a primitive triple. Give the side lengths for each selected choice, and state why each rejected choice fails.
1. Hint 1
  The odd parameter $15$ needs an even partner for opposite parity.
2. Hint 2
  An even partner can still share one of the prime factors of $15$ .
Answer

Only $n=4$ works, producing $(209,120,241)$ . The input $6$ shares $3$ with $15$ , $7$ has the same parity as $15$ , and $10$ shares $5$ with $15$ .

Full solution

Choice $4$ is even and has no common prime factor with $15$ .
Its generated triple is primitive.

Compute the three outputs.
$\begin{aligned}15^2-4^2&=209\\2\cdot15\cdot4&=120\\15^2+4^2&=241\end{aligned}$

Choice $6$ is even but shares the prime $3$ with $15$ .

Choice $7$ is coprime to $15$ but both inputs are odd.

Choice $10$ is even but shares the prime $5$ with $15$ .

Answer

Only $n=4$ works, producing $(209,120,241)$ . The input $6$ shares $3$ with $15$ , $7$ has the same parity as $15$ , and $10$ shares $5$ with $15$ .

Key idea

An input pair must pass both tests before its output is primitive.
Problem 7 Two measurements about the same sides

In a primitive right triangle, let $a$ be the odd leg and $c$ the hypotenuse. Their sum is $578$ , and their difference, longer minus shorter, is $288$ .

Find the parameters $m>n$ and all three side lengths.
1. Hint 1
  The two supplied numbers are twice the parameter squares.
2. Hint 2
  Take positive roots of the halved values and use all three generation formulas.
Answer

The parameters are $(17,12)$ . The odd leg is $145$ , the even leg is $408$ , and the hypotenuse is $433$ .

Full solution

Divide the larger supplied number by $2$ to get the larger parameter square.
$578/2=289$

Divide the smaller supplied number by $2$ to get the smaller parameter square.
$288/2=144$

Their positive roots give
$(m,n)=(17,12).$
The two square values determine the odd leg and hypotenuse, and the roots determine the even leg.

Substitute into the three formulas.
$\begin{aligned}289-144&=145\\2\cdot17\cdot12&=408\\289+144&=433\end{aligned}$

The doubled product labels the horizontal leg, and the square difference labels the vertical leg.

Answer

The parameters are $(17,12)$ . The odd leg is $145$ , the even leg is $408$ , and the hypotenuse is $433$ .

Key idea

The sum and difference of the hypotenuse and odd leg determine both parameter squares.
Problem 8 A square product with shared primes

A right triangle has integer side lengths $525$ , $864$ , and $1{,}011$ . Kai calculates two values from the hypotenuse and the odd leg.

The half-sum is $768$ .

The half-difference is $243$ .

Their product is $432^2$ . Kai concludes that each factor is a square because the product is a square.

Explain the gap in Kai's argument. Find the primitive triple, its two parameters, and the scale factor to the given triangle.
1. Hint 1
  For the separate factors to inherit paired primes, they need an additional property.
2. Hint 2
  Check what happens when you divide every original side by $3$ .
3. Hint 3
  Use the odd leg and hypotenuse of the reduced triangle.
Answer

The factors share $3$ , so the square product does not force each factor to be square. The primitive triple is $(175,288,337)$ , its parameters are $(16,9)$ , and the scale factor is $3$ .

Full solution

The prime factorizations are
$\begin{aligned}768&=2^8\cdot3\\243&=3^5\end{aligned}$

The copies of $3$ pair up in the product because the exponents add to six.
Neither factor has an even exponent of $3$ on its own.

Divide all three sides by $3$ to get $(175,288,337)$ .
The prime factors of $175$ are $5$ and $7$ , neither of which divides $288$ .
The quotient is primitive and $3$ is the full GCF.

Find the half-sum and half-difference using the new hypotenuse and odd leg.
$\begin{aligned}(337+175)/2&=256\\(337-175)/2&=81\end{aligned}$

Their positive roots give
$(m,n)=(16,9).$

Answer

The factors share $3$ , so the square product does not force each factor to be square. The primitive triple is $(175,288,337)$ , its parameters are $(16,9)$ , and the scale factor is $3$ .

Key idea

Remove the full scale factor before using the argument about coprime factors of a square.
Problem 9 A whole family from neighboring integers

Take any positive integer $n$ and use $m=n+1$ in Euclid's formulas.

Prove that the output is primitive. Show that its hypotenuse is one unit longer than its even leg, and give an expression for its odd leg in terms of $n$ .
1. Hint 1
  What would a shared divisor of two consecutive integers have to divide?
2. Hint 2
  To compare the hypotenuse and even leg, subtract their formulas and look for a binomial square.
3. Hint 3
  The odd leg uses the difference of the two parameter squares.
Answer

The output is primitive, its hypotenuse exceeds its even leg by $1$ , and its odd leg is $2n+1$ .

Full solution

A shared divisor of $n$ and $n+1$ must divide their difference, which is $1$ .
The inputs are coprime.
They are consecutive, so exactly one is even.

The two primitive conditions therefore hold for every positive integer $n$ .

Subtract the even-leg expression from the hypotenuse expression.
$\begin{aligned}c-b&=m^2+n^2-2mn\\&=(m-n)^2\\&=1\end{aligned}$

For the odd leg, factor the difference of squares.
$\begin{aligned}a&=(m-n)(m+n)\\&=1\cdot(2n+1)\\&=2n+1\end{aligned}$

Answer

The output is primitive, its hypotenuse exceeds its even leg by $1$ , and its odd leg is $2n+1$ .

Key idea

Consecutive parameters make the square of their difference exactly one.
Problem 10 Prove that the recovery always works

Let a primitive right triangle have positive integer sides and hypotenuse $c$ .

Part A. Prove that exactly one leg is even and that $c$ is odd. Call the odd leg $a$ and the even leg $b$ .

Part B. Let $u$ and $v$ be the half-sum and half-difference of $c$ and $a$ , respectively. Prove they are positive coprime integers whose product is a square.

Part C. Explain why $u$ and $v$ must each be a square, and show how their positive roots give all three side lengths. Justify coprimality and opposite parity of the roots.
1. Hint 1
  For Part A, compare the remainders of even and odd squares after dividing by four.
2. Hint 2
  For Part B, a prime shared by the two half-values would divide their sum and difference.
3. Hint 3
  For Part C, count how often each prime occurs in the product and in the one factor where it appears.
Answer

Part A gives one even leg and an odd hypotenuse. Part B gives positive coprime integers with product $(b/2)^2$ . Part C gives coprime roots of opposite parity with
$\begin{aligned}a&=m^2-n^2\\b&=2mn\\c&=m^2+n^2\end{aligned}$

Full solution

Part A.
Two even legs force an even hypotenuse and a shared factor of $2$ .
Two odd legs make the sum of their squares leave remainder $2$ on division by $4$ , whereas squares leave $0$ or $1$ .

One leg is therefore odd and one is even.
Their square sum is odd, so the hypotenuse is odd.

Part B.
The lengths $c$ and $a$ are both odd and $c>a>0$ , so $u$ and $v$ are positive integers with $u>v$ .

Their product is
$\begin{aligned}uv&=(c+a)(c-a)/4\\&=(c^2-a^2)/4\\&=(b/2)^2\end{aligned}$

A prime shared by $u$ and $v$ would divide their sum $c$ and their difference $a$ .
It would then divide $b^2$ and therefore $b$ , which is impossible for a primitive triple.

Part C.
In the square product every prime occurs an even number of times.
Since each prime occurs in just one factor, its exponent in that factor is even.
The factors are squares.

Let $m$ and $n$ be the positive roots of $u$ and $v$ .
They satisfy $m>n>0$ , share no prime factors, and have opposite parity because their squares add to the odd hypotenuse.

Recover the lengths from the difference, positive square root of the product, and sum.
$\begin{aligned}a&=m^2-n^2\\b&=2mn\\c&=m^2+n^2\end{aligned}$

Answer

Part A gives one even leg and an odd hypotenuse. Part B gives positive coprime integers with product $(b/2)^2$ . Part C gives coprime roots of opposite parity with
$\begin{aligned}a&=m^2-n^2\\b&=2mn\\c&=m^2+n^2\end{aligned}$

Key idea

Coprimality is what lets the square product split into two square factors.

← Back to lesson