I am doing prealgebra 2 and this feels like I just entered another realm of maths

Let $a=7x,b=7y,c=7z,d=7w$, and so we have $x+y+z+w=17^2, 17^2 \mid (xy-zw), 17^2 \mid (x^2+y^2+z^2+w^2)$. Note that if one of $x,y,z,w,$ is a multiple of $17$, let it be $x$, then $17 \mid zw$ and so if WLOG $17 \mid z$ we obtain $17 \mid (y+w)$ and $17 \mid (y^2+w^2),$ from which we infer that $17 \mid y,w$ too, as desired. Now, assume that none of $x,y,z,w$ is a muliple of $17$. Then, $x \equiv \dfrac{zw}{y} \pmod {17^2}$, and since $17 \mid x+y+z+w,x^2+y^2+z^2+w^2$ too, we infer that $17 \mid (y+w)(y+z)$ and $17 \mid (y^2+w^2)(y^2+z^2)$. Note that if $17^2 \mid y+w$ or $17^2 \mid y+z$, then $y+w \geq 17^2=x+y+z+w>y+w,$ or $y+z \geq 17^2=x+y+z+w>y+z,$ a contradiction. Therefore, $17 \mid y+w$ and $17 \mid y+z$. In a similar manner $17 \mid y^2+w^2$ and $17 \mid y^2+z^2$, and so we infer that $17 \mid y,z,w$, a contradiction. Thus, we may conclude. @below yeah oops I mistyped.

Orestis_Lignos wrote: Let $a=17x,b=17y,c=17z,d=17w$, and so we have $x+y+z+w=17^2, 17^2 \mid (xy-zw), 17^2 \mid (x^2+y^2+z^2+w^2)$. Note that if one of $x,y,z,w,$ is a multiple of $17$, let it be $x$, then $17 \mid zw$ and so if WLOG $17 \mid z$ we obtain $17 \mid (y+w)$ and $17 \mid (y^2+w^2),$ from which we infer that $17 \mid y,w$ too, as desired. Now, assume that none of $x,y,z,w$ is a muliple of $17$. Then, $x \equiv \dfrac{zw}{y} \pmod {17^2}$, and since $17 \mid x+y+z+w,x^2+y^2+z^2+w^2$ too, we infer that $17 \mid (y+w)(y+z)$ and $17 \mid (y^2+w^2)(y^2+z^2)$. Note that if $17^2 \mid y+w$ or $17^2 \mid y+z$, then $y+w \geq 17^2=x+y+z+w>y+w,$ or $y+z \geq 17^2=x+y+z+w>y+z,$ a contradiction. Therefore, $17 \mid y+w$ and $17 \mid y+z$. In a similar manner $17 \mid y^2+w^2$ and $17 \mid y^2+z^2$, and so we infer that $17 \mid y,z,w$, a contradiction. Thus, we may conclude. Assuming that each of the numbers $a, b, c, d$ is divisible by $7$, prove that each of the numbers $a, b, c, d$ is divisible by $17$.