The 3 variable square and product

Let $x = y + a$ and $z = y + b$ for some real $a,b.$ The given equations then become

• (i) $(y + a)^{2} - y(y + b) = 37 \Longrightarrow (2a - b)y = 37 - a^{2},$

• (ii) $y^{2} - (y + a)(y + b) = 1 \Longrightarrow -(a + b)y = 1 + ab$ and

• (iii) $(y + b)^{2} - (y + a)y = -35 \Longrightarrow (2b - a)y = -35 - b^{2}.$

Adding these equations together yields that

$0 = 3 - (a^{2} - ab + b^{2}) \Longrightarrow a^{2} - ab + b^{2} = 3,$ (iv).

Now going back to the original equations, when we subtract the second equation from the first we find that

$x^{2} - y^{2} - yz + xz = 37 - 1 \Longrightarrow (x - y)(x + y + z) = 36.$

Now since we are only looking for positive solutions, we must have $x \gt y.$ Similarly, subtracting the third equation from the second, we find that

$y^{2} - z^{2} - xz + xy = 36 \Longrightarrow (y - z)(x + y + z) = 36.$

Again, since we are looking for only positive solutions, we must have $y \gt z.$ Now comparing these last two results, we see that $x - y = y - z,$ implying that $a = -b.$ Plugging this result ito equation (iv) yields that

$a^{2} + a^{2} + a^{2} = 3 \Longrightarrow a^{2} = 1 \Longrightarrow a = 1,$

since $a = x - y \gt 0.$ This in turn gives $b = -1.$ Plugging these results into equation (i) yields that

$3y = 37 - 1 \Longrightarrow y = 12,$ and so $x = 13, z = 11.$

Thus $10000x + 100y + z = 10000*13 + 100*12 + 11 = \boxed{131211}.$

See the second equation which can be written as:
$y^{2} - 1 = xz$
$=> (y+1)(y-1) = xz$
comparing LHS and RHS:
$x = (y+1)$ and $z = (y-1)$
Use this result in any of the eq. 1 or 3 to get a quadratic eq. and solve for y.
eg. In eq.1, we get:
$(y+1)^{2} - y(y-1) = 37$
$=> y = 12$
So,
$x = 13, y = 12, z = 11$

${ \begin{cases} {x^2-yz=37} \\ {y^2-xz=1} \\ {z^2-xy=-35} \end{cases} } \stackrel{\small \mbox{Summing all equations}}{\huge \Longrightarrow} x^2+y^2+z^2-xy-xz-yz=3$

$\Rightarrow 2x^2+2y^2+2z^2-2xy-2xz-2yz=6$

$\Rightarrow (x-y)^2+(x-z)^2+(y-z)^2=6$

Perceive that $y-z = y-x+x-z$ , so let´s do:

${ \begin{cases} {x-y=a} \\ {x-z=b} \\ {y-z=b-a} \end{cases} }$

$\Rightarrow$

${ \begin{cases} {x-y=a \Leftrightarrow x=a+y} \\ {x-z=b \Leftrightarrow z=a-b+y} \\ {y-z=b-a \Leftrightarrow y-(a-b+y)=b-a \Leftrightarrow b-a=b-a} \end{cases} }$

That shows us that $x$ and $z$ are in function of $y$ .

In fact, $y^2-xz=1 \Leftrightarrow xz=y^2-1 \Leftrightarrow xz=(y+1)(y-1)$

If it was $x=y-1, z=y+1$ ,

$(y-1)^2-y(y+1)=37 \Leftrightarrow y^2-2y+1-y^2-y=37$

$\Leftrightarrow -3y=36 \Leftrightarrow y=-12$

This is not possible because $y>0$ .

So,

${ \begin{cases} {x=y+1} \\ {z=y-1} \end{cases} }$

Thus,

$(y+1)^2-y(y-1)=37 \Leftrightarrow y^2+2y+1-y^2+y=37$

$\Leftrightarrow 3y=36 \Leftrightarrow \boxed{y=12} \Leftrightarrow \boxed{x=13} \Leftrightarrow \boxed{z=11}$

Therefore,

$10000x+100y+z=10000.13+100.12+11=\boxed{\boxed{131211}}$

Notes :

$1)$ $a=1, b=2;$

$2)$ $(y-1)^2-y(y+1)^2=-35 \Leftrightarrow y^2-2y+1-y^2-y=-35$

$\Leftrightarrow -3y=-36 \Leftrightarrow y=12;$

$3)$ $y^2-(y+1)(y-1)=1 \Leftrightarrow y^2-y^2+1=1.$

Subtract equation 1 and 2

(x-y)(x+y+z)= 36 (4)

Similarly, subtract equation 2 and 3

(y-z)(x+y+z)=36 (5)

Divide 4 and 5, x-y=y-z, hence x y and z are in AP

x= y-d, z=y+d, (where d =common difference)

So equation 4 becomes, yd= -12. As y is positive, d has to be negative

Substitute the value of x y and z in equation 1 and replace yd by -12, solve to get d =-1.

Hence, x=13, y=12, z=11