a^2+b^2+c^2, a^2+b^2+d^2, a^2+c^2+d^2, b^2+c^2+d^2

 
 
To find positive integers   a,   b,   c,   and   d   such that

a^2 \; + \; b^2 \; + \; c^2 \; = \; M^2
a^2 \; + \; b^2 \; + \; d^2 \; = \; N^2
a^2 \; + \; c^2 \; + \; d^2 \; = \; P^2
b^2 \; + \; c^2 \; + \; d^2 \; = \; Q^2

 
 

Here’s a couple of examples if we let   a = d

a = 120,    b = 28,    c = 21,    d = 120

120^2 \; + \; 28^2 \; + \; 21^2 \; = \; 125^2 \; = \; 5^6
120^2 \; + \; 28^2 \; + \; 120^2 \; = \; 172^2
120^2 \; + \; 21^2 \; + \; 120^2 \; = \; 171^2
28^2 \; + \; 21^2 \; + \; 120^2 \; = \; 125^2 \; = \; 5^6

a = 960,    b = 168,    c = 224,    d = 960

960^2 \; + \; 168^2 \; + \; 224^2 \; = \; 1000^2 \; = \; 10^6
960^2 \; + \; 168^2 \; + \; 960^2 \; = \; 1368^2
960^2 \; + \; 224^2 \; + \; 960^2 \; = \; 1376^2
168^2 \; + \; 224^2 \; + \; 960^2 \; = \; 1000^2 \; = \; 10^6

 
Can you find solutions where   a,   b,   c,   and   d   are all distinct?

 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Advertisements

About benvitalis

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s