Pythagorean Triples Count

Number Theory Level 3

How many pairs of positive integers ( b , c ) (b, c) are there such that

1726 6 2 + b 2 = c 2 ? \large 17266^2+b^2=c^2?

This problem is of this series: Number is all around

∞ \infty 4 1 0 3 2

Md Mehedi Hasan by Md Mehedi Hasan , 22, Bangladesh

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1 solution

Md Mehedi Hasan
Nov 11, 2017

Relevant Wiki: Pythagorean Triple

From Euclid's Formula we write 17266 = 2 k m n b = k ( m 2 − n 2 ) c = k ( m 2 + n 2 ) 17266=2kmn\\b=k(m^2-n^2)\\c=k(m^2+n^2) where k , m , n ϵ N k,m,n\epsilon\mathbb N and m > n m>n

Then we get k m n = 8633 kmn=8633

The divisor of 8633 8633 are 1 , 89 , 97 , 8633 1,89,97,8633 . So k k could be 1 , 89 , 97 1,89,97

1. 1. For k = 1 k=1 ; m n = 8633 mn=8633

If m = 8633 , n = 1 m=8633, n=1 we get b = 863 3 2 − 1 2 = 74528688 b=8633^2-1^2=74528688 and c = 863 3 2 + 1 2 = 74528690 c=8633^2+1^2=74528690

Again m = 97 , n = 89 m=97, n=89 we get b = 9 7 2 − 8 9 2 = 1488 b=97^2-89^2=1488 and c = 9 7 2 + 8 9 2 = 17330 c=97^2+89^2=17330

Here we get 2 \color{#3D99F6}2 triples. They are: ( 17266 , 74528688 , 74528690 ) (17266,74528688,74528690) and ( 17266 , 1488 , 17330 ) (17266,1488,17330)

2. 2. For k = 89 k=89 ; m n = 97 mn=97

If m = 97 , n = 1 m=97,n=1 we get b = 89 ( 9 7 2 − 1 2 ) = 837312 b=89(97^2-1^2)=837312 and c = 89 ( 9 7 2 + 1 ) = 837490 c=89(97^2+1)=837490

Here we get 1 \color{#3D99F6}1 triple. It is ( 17266 , 837312 , 837490 ) (17266,837312,837490)

3. 3. For k = 97 k=97 ; m n = 89 mn=89

If m = 89 , n = 1 m=89,n=1 we get b = 97 ( 8 9 2 − 1 2 ) = 768240 b=97(89^2-1^2)=768240 and c = 97 ( 8 9 2 + 1 2 ) = 768434 c=97(89^2+1^2)=768434

Here we get 1 \color{#3D99F6}1 triple. It is ( 17266 , 768240 , 768434 ) (17266,768240,768434)

So total triples are 4 \color{#3D99F6}\boxed{4}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

What about ( 17266 , 768240 , 768434 ) (17266, 768240,768434) ?

Munem Shahriar - 3 years, 6 months ago

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Thank you, bro. You are right. I didn't notice it before. Now I have updated my solution and reported in report section to update the answer.

Md Mehedi Hasan - 3 years, 6 months ago

Another possible way to solve it would be to just subtract b 2 b^2 onto the other side; then factor the right side as ( c + b ) ( c − b ) (c+b)(c-b) . Then using the fact that the prime factorization of 17266 17266 is 2 ⋅ 89 ⋅ 97 2\cdot 89\cdot 97 , set up all the possible systems of equations of the form c + b = x , c − b = y c+b=x,c-b=y , where x , y , b , c x,y,b,c are integers, x y = 1726 6 2 xy=17266^2 , and x > y x>y ( x > y x>y because b b and c c are positive). You can cut the number of combinations by noting that x x and y y must be even because otherwise b b and c c would not be integers. Skipping the process of trying all such combinations, the four systems are c + b = 2 ⋅ 9 7 2 , c − b = 2 ⋅ 8 9 2 c+b=2\cdot 97^2, c-b=2\cdot 89^2 c + b = 2 ⋅ 8 9 2 ⋅ 9 7 2 , c − b = 2 c+b=2\cdot 89^2\cdot 97^2,c-b=2 c + b = 2 ⋅ 8 9 2 ⋅ 97 , c − b = 2 ⋅ 89 c+b=2\cdot 89^2\cdot 97, c-b=2\cdot 89 c + b = 2 ⋅ 89 ⋅ 9 7 2 , c − b = 2 ⋅ 89 c+b=2\cdot 89\cdot 97^2, c-b=2\cdot 89

James Wilson - 3 years, 5 months ago

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