MHB Amazing Math Discovery: 8281 is the Only 4 Digit Square with this Quirk

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A unique four-digit number, n, exists where the two-digit number formed by the first two digits is one greater than the two-digit number formed by the last two digits, and n is a perfect square. This number is 8281. The discussion includes a challenge to prove its uniqueness without relying on extensive trial and error. Participants engage in a light-hearted exchange about the proof, with one asserting that the proof is intuitively obvious and another offering a mathematical expression to support the claim. The conversation highlights the mathematical curiosity surrounding this specific property of the number 8281.
Wilmer
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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 :(
 
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Wilmer said:
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
 
I found it...you prove it...why should I have all the fun!
 
Wilmer said:
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
 
topsquark said:
I can prove it. It's intuitively obvious. (Quote from my Stat Mech professor.)

-Dan
Proof below
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 $91^2=8281$
 
No...no...
a = 1st digit, b = 2nd digit:

1010a + 101b - 1 = x^2 :)
 
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