square numbers sequence

let $N_{m}= \underbrace{aa \cdots a}_{\text{m times}} \underbrace{bb \cdots b}_{\text{m-1 times}}c$

Then, using that $\underbrace{11 \cdots 1}_{\text{y times}}= \frac{10^{y}-1}{9}$

$N_{m}= a\times 10 ^{m} \times \frac{10^{m}-1}{9} + b\times 10\times \frac{10 ^{m-1}-1}{9} + c$

If $x=10 ^{m}$

$N_{m}=\frac{x^2-x}{9}\times a+ \frac{x-10}{9}\times b + c \\$

We will assume that $N_{m}=u^2$ with $u$ an integer $\\$

$\begin{aligned} u^2=\frac{x^2-x}{9}\times a+ \frac{x-10}{9}\times b + c \\(3u)^2= (a)x^2+(b-a)x+9c-10b \end{aligned} \\$

The above equation is cuadratic, and if it is a perfect square, then, the discriminant is equal to 0

$\begin{aligned} (b-a)^2 - 4(a)(9c-10b)=0 \\ b^2-2ab+a^2-36ac+40ab=0 \\ b^2+2ab+a^2-36ac+40ab=4ab \\ (a+b)^2=36ac-36ab \\ (a+b)^2= 36a(c-b) \end{aligned}$

We get that

$6 \vert {(a+b)} , b<c$ and $a \vert b$

Trying values for $a$ give us all the triplets: $(1,5,9)$ and $(4,8,9)$

Just searched for cases ac,aabc using $Mathematica$

Select[Tuples[Range@9,{3}],IntegerQ@Sqrt[FromDigits@{#,#3}&@@#]&&IntegerQ@Sqrt[FromDigits@{#,#,#2,#3}&@@#]&]

and got the right answer

{{1, 5, 6}, {4, 8, 9}}