When a*b*x*y and (a^2 + b^2)(x^2 + y^2) are both squares

 
 
a \cdot b \cdot x \cdot y \; = \; N^2
(a^2 \; + \; b^2) \,(x^2 \; + \; y^2) \; = \; M^2

 
 
Here’s one possible solution:

(15,   112,   113),   a Primitive Pythagorean triple, and
(12,   35,   37),   also a Primitive Pythagorean triple.

a = 15,   b = 112,      x = 12,   y = 35

(a^2 \; + \; b^2) \,(x^2 \; + \; y^2)

15^2 \; + \; 112^2 \; = \; 113^2 \; .......... \; 12^2 \; + \; 35^2 \; = \; 37^2

113^2 \; \times \; 37^2 \; = \; 4181^2

a \cdot b \cdot x \cdot y \; = \; 15 \times 112 \times 12 \times 35 \; = \; 705600 \; = \; 840^2

 
 
Find other solutions.

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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