Taking x and y as the integers.
Let's say that x is 4 less than twice y. This is:
[tex]x=2y-4[/tex]The sum of the squares of x and y is 544, this is:
[tex]x^2+y^2=544[/tex]Use the first equation and replace this expression for the value of x in the second equation:
[tex]\begin{gathered} (2y-4)^2+y^2=544 \\ 4y^2-16y+16+y^2=544 \\ 5y^2-16y+16-544=0 \\ 5y^2-16y-528=0 \end{gathered}[/tex]Solve the quadratic equation
[tex]\begin{gathered} y=\frac{16\pm\sqrt[]{16^2-4(5\cdot-528)}}{2\cdot5} \\ y=\frac{16\pm104}{10} \\ y1=\frac{16+104}{10}=\frac{120}{10}=12 \\ y2=\frac{16-104}{10}=-\frac{44}{5} \end{gathered}[/tex]y can take 2 values, but, as we know, x and y must be positive integers, the value of y that meets this condition is 12. It means one of the integer is 12.
Use this value and the first equation to find the second integer:
[tex]\begin{gathered} x=2y-4 \\ x=2\cdot12-4 \\ x=24-4 \\ x=20 \end{gathered}[/tex]The integers are 20 and 12.
