MHB Can You Solve These Pythagorean Quadruples?

  • Thread starter Thread starter Albert1
  • Start date Start date
AI Thread Summary
The discussion presents three Pythagorean quadruple equations requiring the identification of natural numbers a and b. Participants are attempting to solve the equations: (1) 12² + 39² + a² = b², (2) 24² + 36² + a² = b², and (3) 15² + 9² + a² = b². Multiple users are engaging with the problems, providing their attempts and solutions. The focus remains on finding valid pairs of natural numbers that satisfy each equation. The discussion emphasizes collaborative problem-solving in mathematics.
Albert1
Messages
1,221
Reaction score
0
(1)$12^2+39^2+a^2=b^2$ find $a,b$
$a,b\in N$
(2)$24^2+36^2+a^2=b^2$ find $a,b$
$a,b\in N$
(3)$15^2+9^2+a^2=b^2$ find $a,b$
$a,b\in N$
 
Mathematics news on Phys.org
Albert said:
(1)$12^2+39^2+a^2=b^2$ find $a,b$
$a,b\in N$
(2)$24^2+36^2+a^2=b^2$ find $a,b$
$a,b\in N$
(3)$15^2+9^2+a^2=b^2$ find $a,b$
$a,b\in N$

let me attempt (3)
$b^2-a^2 = 15^2 + 9^2 = 306$ this is of the form 4n +2 so one factor is odd and another even so no solution

For (2)

$b^2-a^2 = 24^2+36^2 = 1872$ so (b-a) and (b+a) both should be even

the factor 2 * 936 giving b = 469 and a = 467
the factor 4 * 468 giving b = 236 and a = 232
the factor 6 * 312 giving b = 159 and a = 153
the factor 8 * 234 giving b = 121 and a = 113
the factor 12 * 156 giving b = 84 and a = 72
the factor 18 * 104 giving b = 61 and a = 43
the factor 24 * 78 giving b = 51 and a = 27
the factor 26 * 72 giving b = 49 and a = 23
the factor 36 * 52 giving b = 44 and a = 8
 
Last edited:
kaliprasad said:
let me attempt (3)
$b^2-a^2 = 15^2 + 9^2 = 306$ this is of the form 4n +2 so one factor is odd and another even so no solution

For (2)

$b^2-a^2 = 24^2+36^2 = 1872$ so (b-a) and (b+a) both should be even

the factor 2 * 936 giving b = 469 and a = 467
the factor 4 * 468 giving b = 236 and a = 232
the factor 6 * 312 giving b = 159 and a = 153
the factor 8 * 234 giving b = 121 and a = 113
the factor 12 * 156 giving b = 84 and a = 72
the factor 18 * 104 giving b = 61 and a = 43
the factor 24 * 78 giving b = 51 and a = 27
the factor 26 * 72 giving b = 49 and a = 23
the factor 36 * 52 giving b = 44 and a = 8

Solution for 1

as the value is 1665 we get factors and then a and b as

the factor 1 * 1665 giving b = 833 and a = 832
the factor 3 * 555 giving b = 279 and a = 276
the factor 5 * 333 giving b = 169 and a = 164
the factor 9 * 185 giving b = 97 and a = 88
the factor 15 * 111 giving b = 63 and a = 48
the factor 37 * 45 giving b = 41 and a = 4
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top