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

  • Context: MHB 
  • Thread starter Thread starter Wilmer
  • Start date Start date
  • Tags Tags
    Discovery
Click For Summary

Discussion Overview

The discussion revolves around a mathematical property of the four-digit number 8281, specifically focusing on its uniqueness as a square number that meets a certain condition regarding the relationship between its first two digits and its last two digits. The scope includes mathematical reasoning and exploration of proofs related to this property.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that 8281 is the only four-digit square number 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.
  • One participant challenges others to prove the uniqueness of this property without relying on extensive trial and error.
  • Another participant expresses a willingness to prove the claim, suggesting that the proof is intuitively obvious.
  • A mathematical expression is presented, indicating a potential approach to proving the uniqueness of 8281 as a square number under the discussed condition.

Areas of Agreement / Disagreement

Participants generally agree on the assertion that 8281 is unique under the specified conditions, but there is no consensus on the method of proof or the validity of the intuitive claims made.

Contextual Notes

The discussion includes assumptions about the properties of square numbers and the specific digits involved, but these assumptions are not fully explored or resolved within the thread.

Wilmer
Messages
303
Reaction score
0
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 :(
 
Mathematics news on Phys.org
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 :)
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 125 ·
5
Replies
125
Views
20K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
2K