$a^2+b^2=ab$

Rearranging we get: $a(10-a)=b(b-1)$ .

RHS is even, hence $a$ must be even.

Due to the symmetry of the LHS, it can only equal 2 values: $2*8=16$ and $4*6=24$ , neither of which can be written in the form $b(b-1)$ .

From $a^2 + b^2 = 10a + b$ , we get $a^2 - 10a + (b^2 - b) = 0$ . Solving the quadratic for $a$ , we have $a = 5 \pm \sqrt {25-b(b-1)}$ . For $a$ to be a positive integer, $25-b(b-1)$ must be positive and a perfect square. For $25-b(b-1) \ge 0$ , $b < 6$ or $b = 0, 1, 2, 3, 4, 5$ . Of all the six possible values of $b$ , only when $b=1$ is $25-b(b-1) = 25$ a perfect square. Then $a = 5 \pm 5 = 0$ or $10$ . Both are unacceptable values for $a$ . Therefore there is no two-digit integer $\overline{ab}$ which satisfies the equation.

If we start with $a^2 + b^2 = 10a + b$ , then we obtain the quadratic equation:

$a^2 - 10a + (b^2 - b) = 0 \Rightarrow a = \frac{10 \pm \sqrt{100 - 4(1)(b^2 -b)}}{2} = \frac{10 \pm \sqrt{-4b^2 + 4b + 100}}{2}$ (i)

We require the discriminant in (i) to be non-negative AND for $b \in \mathbb{Z_{0}^{+}}$ , which gives us:

$-4b^2 + 4b + 100 \ge 0 \Rightarrow b = \frac{-4 \pm \sqrt{16 - 4(-4)(100)}}{-8} = \frac{1 \pm \sqrt{101}}{2} \Rightarrow 0 \le b \le 5$ ;

or $b = [0,1,2,3,4,5]$ . Substituting each of these values back into (i) gives the ordered pairs:

$(a,b) = (0,0); (10,0); (0,1); (10,1); (5+\sqrt{23},2); (5-\sqrt{23},2); (5+\sqrt{19},3); (5-\sqrt{19},3); (5+\sqrt{13},4); (5-\sqrt{13},4); (5+\sqrt{5},5); (5-\sqrt{5},5).$

There are NO positive integral values for $a$ to take on under these conditions, which means there is no two-digit number $ab$ that satisfies the original Diophantine equation above.

After some manipulation, the equation becomes $(2a-10)^2 + (2b-1)^2 = 101.$ The only way to write $101$ as a sum of squares, up to sign and rearrangement, is $10^2+1^2.$ (This is easy to check by inspection, or using the fact that $101$ is prime and knowing some number theory.)

Since $2a-10$ is even, it must be $\pm 10,$ which gives $a = 0, 10.$ Neither of these can be the first digit of a two-digit number. So the answer is No.

(The complete set of solutions $(a,b)$ to the given equation is $\{(0,0), (0,1), (10, 0), (10, 1)\}.)$