4-digit square where first two digits exceed last two digits by 1

[Wilmer]

Just discovered that:
if n is a 4 digit number such that the 2 digit number
formed by its 1st 2 digits is 1 greater than the
2 digit number formed by its last 2 digits,
and n is a square, then only one such n exists: 8281

Unfortunately, such a mathshaking discovery
will not bring down the price of groceries :(

 

[topsquark]

Just discovered that:
if n is a 4 digit number such that the 2 digit number
formed by its 1st 2 digits is 1 greater than the
2 digit number formed by its last 2 digits,
and n is a square, then only one such n exists: 8281

Unfortunately, such a mathshaking discovery
will not bring down the price of groceries :(

Okay, so prove it's unique! And no making a big list of trial and error or you'll get coal in your stocking.

-Dan

 

[Wilmer]

I found it...you prove it...why should I have all the fun!

 

[topsquark]

I found it...you prove it...why should I have all the fun!

I can prove it. It's intuitively obvious. (Quote from my Stat Mech professor.)

-Dan

 

[kaliprasad]

I can prove it. It's intuitively obvious. (Quote from my Stat Mech professor.)

-Dan

Proof below

Spoiler

let the 2nd 2 digits be n so 1st 2 digits (n+1) and number be $x^2$

$100(n+1) + n = 101n + 100 = x^2$

now $101n = x^2 - 100 = (x+10)(x-10)$ and as 101 > n and 101 is prime so

x + 10 = 101k as as x is 2 digit number k =1 so x = 91 and we get the number $91^2=8281$

 

[Wilmer]

No...no...
a = 1st digit, b = 2nd digit:

1010a + 101b - 1 = x^2 :)