Find two possible 2-digit numbers that satisfy the given conditions

Question

Find two possible solutions to a 2-digit number equal to the square of the ten's digit plus the square of the sum of its digits.

I managed to find out through brute force that the two solutions that satisfy the question is 41 and 50. However, I now need to provide an explanation for how I got my answer.

How would I solve this type of question in a formal manner?

Answer 1

Hint:

$$10a+b=a^2+(a+b)^2\iff2a^2+2a(b-5)+b^2-b=0$$

Discriminant $$=4(b-5)^2-8(b^2-b)=-4b^2-32b+100=4(25-16b-b^2)=4(89-(b+8)^2)$$ has to be perfect sqaure $=(2c)^2$(say)

$$89=c^2+(b+8)^2$$

As $b\ge0,b+8\ge8$

Again, $(b+8)^2=89-c^2\le89\implies b+8\le\sqrt{89}<10$

Answer 2

Let the digits be $a$ and $b$. Solving for $a$ in $10a+b=a^2+(a+b)^2$ using the quadratic formula, we have:

$$a=\frac{1}{2} \left(-\sqrt{-b^2-8b+25}-b+5 \right)$$ $$a=\frac{1}{2} \left(\sqrt{-b^2-8b+25}-b+5 \right)$$

For $b=0$, $b=1$ and $b=2$, there are solutions. For $b =3$ the part under the square root is negative, so there is no solution.

And since the vertex of $-b^2-8b+25$ has $x$-coordinate $-4$, and the quadratic is concave down, the function inside the square root is decreasing. Thus there are no solutions when $b > 3$.