Question 867346
let the odd integers be x, x+2,x+4

sum of squares of first two
x^2+(x+2)^2

which is equal to

(x+4)^2+9

x^2+(x+2)^2=(x+4)^2+9

x^2+x^2+4x+4=x^2+8x+16+9

2x^2+4x+4=x^2+8x+25


x^2-4x-21=0

x^2-7x+4x-21=0

x(x-7)+4(x-7)=0

(x-7)(x+4)=0

x=7 Or -4

but the number is positive
so the numbers are 7,9,11