Determine the numbers a and b that check the equality $\sqrt{\overline{aba}}=(a+b-1)\cdot \sqrt{a+b}$

Question

Determine the numbers a and b that check the equality

$\sqrt{\overline{aba}}=(a+b-1)\cdot \sqrt{a+b}$

MY IDEAS

I thought of decomposing $\overline{aba}$ as $101\cdot a + 10\cdot b$

Then i thought that the radicals are making the equation harder so I raised everything to the power of 2.

$101\cdot a + 10\cdot b= {(a+b-1)}^{2}\cdot (a+b)$

OTHER THINGS I THOUGHT OFF;

What should I donext? I tried descomposing the ${(a+b-1)}^{2}$, but didn't get anything helpful. Any idea is welcome! Thank you!!

Answer 1

You have got to this point:

$$\overline{aba}=(a+b-1)^2(a+b)$$

Now note that the right side only depends on $a+b$, and consider the possible values of $a+b$. You get $6\le a+b\le 10$ (as for $a+b\le 5$ is too small and $a+b\ge 11$ is too big to get a $3$-digit number on the right side).

For the values of $a+b$ between $6$ and $10$ you get the following table:

$$\begin{array}{c|c}a+b&(a+b-1)^2(a+b)\\\hline6&150\\7&252\\8&392\\9&576\\10&810\end{array}$$

Nice. So only for $a+b=7$ the number is of the form $\overline{aba}$. Also, as it happens, with $a=2, b=5$ we do have $a+b=7$. So $a=2, b=5$ is the unique solution.

Answer 2

I assume that you ask about $a,b$ integers because if not for each real $a$ you have a real $b$ for which $$101\cdot a + 10\cdot b= {(a+b-1)}^{2}\cdot (a+b)$$ We deduce the equation $$\frac{91a}{a+b}+10=(a+b-1)^2$$ It follows the possible cases are $$a+b=91a,7a,13a,7,13$$ and the only valid of them is $a+b=7$ from which the only solution $(a,b)=(2,5)$.