A number theory problem by Priyanshu Mishra

$\sqrt [ 3 ]{ 7{ x }^{ 2 } - 13xy + 7{ y }^{ 2 } } = |x - y| + 1$

Let $x \geq y$ . Cubing both sides

$7{ x }^{ 2 } - 13xy + 7{ y }^{ 2 }=(x-y+1)^3$

$7{ x }^{ 2 } - 13xy + 7{ y }^{ 2 }=(x-y)^3 +1+ 3(x-y).1(x-y+1)$

$7{ x }^{ 2 } - 14xy + 7{ y }^{ 2 }=(x-y)^3 +1 +3 (x-y)(x-y+1) -x y$

$7 (x-y)^2=(x-y)^3 +1 +3 (x-y)(x-y+1) -x y$

$7 (x-y)^2=(x-y)^3 +1 +3 (x-y)^2+3(x-y) -x y$

$(x-y)^3-4(x-y)^2+3 (x-y)=x y-1$

Factoring L.H.S.

$(x-y)(x-y-1)(x-y-3)=x y-1$

Let consider $x-y=z \implies y=x-z$ .putting this in this equation

$z(z-1)(z-3)=x (x-z) -1 \\ x^2 - z.x- (1+z(z-1)(z-3))=0$

As $x$ is a positive integer so its discriminant will be a perfect square

$D= \sqrt{z^2+4+4z(z-1)(z-3)}$ .For smallest solution put $z=0$ .It will give $x-y=0 \implies x=y$ .

Putting this in the given question we get $x=y=1$

For the second case $x \leq y$ will arise same solution

So,the smallest solution is $(x,y)=(1,1)$