Determine numbers for which $\sqrt{ \frac{\overline{ab}^2 + \overline{cd}^2}{2}} = \frac{ \overline{ab} + \overline{cd}}{ 2} +1$.

Question

The question

Determine the pairs of numbers of the form $\overline{ab} , \overline{cd}$ (numbers with 2 digits in base 10) for which the equality $$\sqrt{ \frac{\overline{ab}^2 + \overline{cd}^2}{2}} = \frac{ \overline{ab} + \overline{cd}}{ 2} +1$$, is true.

My idea

I noted the numbers as x and y to be more easy to write.

If we square it up and then multiply by 4 we get that:

$2x^2+2y^2=x^2+y^2+2xy+4x+4y+4$

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

From here we get that $x+y+1$ is a square number.

I thought of using modular arithmetic.

I dont know what to do forward. Thank you!

Answer 1

From the equation,
$x^2-2xy+y^2-4x-4y-4=0$,
We try to solve for y in terms of x using the quadratic formula. We obtain
$y = x + 2\pm\sqrt{8(x+1)}$, or
$y = x + 2\pm2\sqrt{2}\sqrt{x+1}$
For y to be integral, $x+1$ must be a multiple of $2$, and $\frac{x+1}{2}$ must be a perfect square.
i.e, $x = 2n^2-1$, then
$y =2n^2+1\pm4n$.

The the possible pairs of numbers with the given condition are: $(17,31),(17,7),(31,49),(31,17),(49,71),(49,31),(71,97),(71,49),(97,71)$

Answer 2

Step 1: Substitute ab and cd for x and y and being double digits means $x,y \in (-100,-10] \cup [10,100)$ and $x,y \in \mathbb{Z}$

Step 2: Solve for x $$\sqrt\frac{x^2+y^2}{2}=\frac{x+y}{2}+1$$ Note that this requires $x+y\geq-2$

Now we can square both sides and multiply by 4 to get to the point you got $$2x^2+2y^2=x^2+2xy+y^2+4x+4y+4$$ Now we can move everything to the left-hand side $$x^2-(2y+4)x+y^2-4y-4=0$$ which is a quadratic equation with a determinant $$D = (2y+4)^2-4(y^2-4y-4)=4y^2+16y+16-4y^2+16y+16 = 16(2y+2)$$ Note that now we need $y\geq-1$ and then we get our solution for x $$x_{1,2}=y+2 \pm 2\sqrt{2y+2} $$ It can be easily seen that this would make x positive as well meaning $x,y\in[10,100)$

And you will get the following pairs $(31,17),(49,31),(71,49),(97,71)$

Obviously x and y are interchangeable so if $(x,y)$ is a solution so is $(y,x)$