MHB Can You Solve These Pythagorean Quadruples?

[Albert1]

(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 $a,b$
$a,b\in N$

 

[kaliprasad]

(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)

Spoiler

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

For (2)

Spoiler

$b^2-a^2 = 24^2+36^2 = 1872$ 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

 

[kaliprasad]

let me attempt (3)

Spoiler

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

For (2)

Spoiler

$b^2-a^2 = 24^2+36^2 = 1872$ 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

Spoiler

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