Question

show that if x and y whole numbers, then the expression x^2+2(x+y)^2+3(x+2y)^2+4(x+3y)^2 can be written as a sum of two square numbers

Answer 1

x^2+2(x^2+2xy+y^2)+3(x^2+4xy+4y^2)+4(x^2+6xy+9y^2)=
x^2+2x^2+4xy+2y^2+3x^2+12xy+12y^2+4x^2+24xy+36y^2
No arithmetic mistakes so far I hope..

Answer 2

I'm skipping some steps in this simplification I'm assuming you can do the necessary steps. \[x^2+2(x+y)^2+3(x+2y)^2+4(x+3y)^2=10x^2+40xy+50y^2\]\[=10(x+2y)^2+10y^2=10((x+2y)^2+y^2)\]\[=(1^2+9^2)((x+2y)^2+y^2)\]Since it's a proven fact that the product of two numbers that are a sum of two squares, can be written as a sum of two squares, we are done.

Answer 3

3x^2+4xy+2y^2+3x^2+12xy+12y^2+4x^2+24xy+36y^2=
10x^2+4xy+2y^2+12xy+12y^2+4x^2+... forget it, KG is too good lol.

Answer 4

Thank You Guys!

Answer 5

Namely, \[((x+2y)+(9y)^2)+(y^2-(9(x+2y))^2)\]

Answer 6

Correction: \[((x+2y)^2+(9y)^2)+(y^2-(9(x+2y))^2)\]

Answer 7

Everywhere I wrote a 9, replace with a 3 because I make stupid typos like that.

Answer 8

I see

Answer 9

I really need to sleep. I'm making typos all over the place. You should be getting \[((x+2y)+(3y))^2+(y-(3(x+2y)))^2=(x+5y)^2+(-3x-5y)^2\]

Answer 10

I apologize for the confusion.