Find A in abcd=A(four digital number) is a perfect square ,given ab=2cd+1

[Albert1]

$\overline{abcd}=A$(four digital nmber) is a perfect square ,given $\overline{ab}=2\overline{cd}+1$
find $A=?$

 

[MarkFL]

My solution:

Spoiler

We may state:

\(\displaystyle (20c+2d+1)100+10c+d=n^2\)

or:

\(\displaystyle 67(30c+3d)=(n+10)(n-10)\)

Let's let $n=77$ so we have:

\(\displaystyle 67(30c+3d)=87\cdot67\)

Hence:

\(\displaystyle 30c+3d=87\)

\(\displaystyle 10c+d=29=10\cdot2+9\)

Thus we see we have:

\(\displaystyle c=2,\,d=9\implies a=5,\,b=9\)

And so:

\(\displaystyle A=5929=77^2\)

 

[I like Serena]

My solution:

Spoiler

Let $x=\overline{ab}$, $y=\overline{cd}$, and $A=n^2$.

Then:
$$A=100x+y = n^2 \land x=2y+1 \quad\Rightarrow\quad
100(2y+1)+y=n^2 \quad\Rightarrow\quad
3\cdot 67 \cdot y=(n-10)(n+10)
$$
Since $67$ is prime, it must divide either $n-10$ or $n+10$.
Since $n$ must be a 2-digit number, we conclude:
$$n-10=67 \lor n+10 = 67 \quad\Rightarrow\quad n=57 \land n=77$$
Only $A=77^2=5929$ satisfies the condition, so that is the one and only solution. (Smile)