Squares and Square Roots | IT

Question 26

Is (5,12,13)(5, 12, 13) a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 2020?

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Solution
Understand the Question
  • A Baudhayana triple (or Pythagorean triple) is a set of three positive integers (a,b,c)(a, b, c) satisfying the relation a2+b2=c2a^2 + b^2 = c^2.
  • A triple is primitive if aa, bb, and cc share no common factor other than 11, meaning HCF(a,b,c)=1\text{HCF}(a, b, c) = 1.
  • Primitive triples are generated by the formulas a=m2n2a = m^2 - n^2, b=2mnb = 2mn, and c=m2+n2c = m^2 + n^2, where m>n>0m > n > 0, HCF(m,n)=1\text{HCF}(m, n) = 1, and one of m,nm, n is even while the other is odd.

(i) Is (5,12,13)(5, 12, 13) a primitive Baudhayana triple?

Step 1 · Verify the Baudhayana Relation

For (5,12,13)(5, 12, 13), let a=5a = 5, b=12b = 12, and c=13c = 13.

Calculate a2+b2a^2 + b^2:

a2+b2=52+122=25+144=169\begin{aligned} a^2 + b^2 &= 5^2 + 12^2 \\ &= 25 + 144 \\ &= 169 \end{aligned}

Calculate c2c^2:

c2=132=169\begin{aligned} c^2 &= 13^2 \\ &= 169 \end{aligned}

Since a2+b2=c2=169a^2 + b^2 = c^2 = 169, (5,12,13)(5, 12, 13) is a Baudhayana triple.

Step 2 · Check for Primitivity

Find the factors of each number:

  • Factors of 55: 1,51, 5
  • Factors of 1212: 1,2,3,4,6,121, 2, 3, 4, 6, 12
  • Factors of 1313: 1,131, 13

The only common factor of 5,12,5, 12, and 1313 is 11, so HCF(5,12,13)=1\text{HCF}(5, 12, 13) = 1.

Therefore, (5,12,13)(5, 12, 13) is a primitive Baudhayana triple.

Answer

(i) Yes, (5,12,13)(5, 12, 13) is a primitive Baudhayana triple.

(ii) What are the other primitive Baudhayana triples with numbers less than or equal to 2020?

Step 1 · Generate Primitive Triples with Values 20\le 20

Primitive Baudhayana triples (a,b,c)(a, b, c) can be generated using:

a=m2n2b=2mnc=m2+n2\begin{aligned} a &= m^2 - n^2 \\ b &= 2mn \\ c &= m^2 + n^2 \end{aligned}

where m>nm > n, HCF(m,n)=1\text{HCF}(m, n) = 1, and one of m,nm, n is even while the other is odd.

Case 1: Let m=2,n=1m = 2, n = 1

a=m2n2=2212=41=3b=2mn=2×2×1=4c=m2+n2=22+12=4+1=5\begin{aligned} a &= m^2 - n^2 = 2^2 - 1^2 = 4 - 1 = 3 \\ b &= 2mn = 2 \times 2 \times 1 = 4 \\ c &= m^2 + n^2 = 2^2 + 1^2 = 4 + 1 = 5 \end{aligned}

Triple: (3,4,5)(3, 4, 5) (all values 20\le 20).

Case 2: Let m=3,n=2m = 3, n = 2

a=m2n2=3222=94=5b=2mn=2×3×2=12c=m2+n2=32+22=9+4=13\begin{aligned} a &= m^2 - n^2 = 3^2 - 2^2 = 9 - 4 = 5 \\ b &= 2mn = 2 \times 3 \times 2 = 12 \\ c &= m^2 + n^2 = 3^2 + 2^2 = 9 + 4 = 13 \end{aligned}

Triple: (5,12,13)(5, 12, 13) (the given triple).

Case 3: Let m=4,n=1m = 4, n = 1

a=m2n2=4212=161=15b=2mn=2×4×1=8c=m2+n2=42+12=16+1=17\begin{aligned} a &= m^2 - n^2 = 4^2 - 1^2 = 16 - 1 = 15 \\ b &= 2mn = 2 \times 4 \times 1 = 8 \\ c &= m^2 + n^2 = 4^2 + 1^2 = 16 + 1 = 17 \end{aligned}

Triple: (15,8,17)(15, 8, 17) or (8,15,17)(8, 15, 17) (all values 20\le 20).

Case 4: Let m=4,n=3m = 4, n = 3

a=m2n2=4232=169=7b=2mn=2×4×3=24c=m2+n2=42+32=16+9=25\begin{aligned} a &= m^2 - n^2 = 4^2 - 3^2 = 16 - 9 = 7 \\ b &= 2mn = 2 \times 4 \times 3 = 24 \\ c &= m^2 + n^2 = 4^2 + 3^2 = 16 + 9 = 25 \end{aligned}

Here b=24>20b = 24 > 20 and c=25>20c = 25 > 20, so this triple is excluded.

For any m5m \ge 5, c=m2+n252+12=26>20c = m^2 + n^2 \ge 5^2 + 1^2 = 26 > 20.

Excluding the given triple (5,12,13)(5, 12, 13), the other primitive triples with numbers 20\le 20 are (3,4,5)(3, 4, 5) and (8,15,17)(8, 15, 17).

Answer

(ii) (3,4,5)(3, 4, 5) and (8,15,17)(8, 15, 17)

Common Mistakes
  • Primitive vs. Non-primitive: Mistaking multiples like (6,8,10)(6, 8, 10) or (9,12,15)(9, 12, 15) for primitive triples. Although they satisfy a2+b2=c2a^2 + b^2 = c^2, they are not primitive because HCF(a,b,c)>1\text{HCF}(a, b, c) > 1.
  • Including the Given Triple: Listing (5,12,13)(5, 12, 13) again in part (ii) despite the question asking for the other primitive triples.
  • Exceeding the Bound: Forgetting the condition that all three numbers must be 20\le 20, which excludes triples like (7,24,25)(7, 24, 25).

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Q20

List down all the Baudhāyana triples with numbers less than or equal to 20.

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Q22

Is (30,40,50)(30, 40, 50) a Baudhāyana triple?

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Q23

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Q24

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Q25

Context: Conjecture: (3k,4k,5k)(3k, 4k, 5k) is a Baudhāyana triple, where kk is any positive integer.

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Q26

Is (5,12,13)(5, 12, 13) a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 2020?

Q27

Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

Q28

If (a,b,c)(a, b, c) is non-primitive, and the integers have ff — greater than 1 — as a common factor, then is (af,bf,cf)\left(\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}\right) a Baudhayana triple? Check this statement for (9,12,15)(9, 12, 15). Justify this statement.

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