Sum of 3 Squares and 4 Square

Number Theory Level pending

It's not original question, so I don't know whether this problem have been posted or not.

Suppose all variables below are integers and relatively prime. How many unordered possible solution for following equation are there?

3 ( x 2 + y 2 + z 2 ) = a 2 + b 2 + c 2 + d 2 3(x^2+y^2+z^2)=a^2+b^2+c^2+d^2

Less than 20 Between 20 and 100 (inclusive) More than 100 but finitely many Infinitely many

Lingga Musroji by Lingga Musroji , 27, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Lingga Musroji
Jan 5, 2021

3 ( x 2 + y 2 + z 2 ) = ( x + y + z ) 2 + ( x 2 − 2 x y + y 2 ) + ( x 2 − 2 x z + z 2 ) + ( y 2 − 2 y z + z 2 ) = ( x + y + z ) 2 + ( x − y ) 2 + ( x − z ) 2 + ( y − z ) 2 3(x^2+y^2+z^2)=(x+y+z)^2+(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)=(x+y+z)^2+(x-y)^2+(x-z)^2+(y-z)^2

By choosing a = x + y + z , b = x − y , c = x − z , d = y − z a=x+y+z,b=x-y,c=x-z,d=y-z , hence there is infinitely many solutions

1 Helpful 0 Interesting 0 Brilliant 0 Confused

It's easy to generate infinitely many solutions from just one; if ( x , y , z , a , b , c , d ) (x,y,z,a,b,c,d) is a solution then so is ( t x , t y , t z , t a , t b , t c , t d ) (tx,ty,tz,ta,tb,tc,td) for any integer t t . Since ( 3 , 2 , 1 , 6 , 2 , 1 , 1 ) (3,2,1,6,2,1,1) is a solution, there are infinitely many solutions.

You've shown more here, though: the solutions can have ( x , y , z , a , b , c , d ) (x,y,z,a,b,c,d) relatively prime, which is definitely more surprising.

Chris Lewis - 5 months, 1 week ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...