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

  • Thread starter Thread starter Wilmer
  • Start date Start date
  • Tags Tags
    Discovery
AI Thread Summary
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
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 :(
 
Physics 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 :)
 
"In 2013, after a series of hospitalizations due to magnets, New Zealand government officials permanently banned the sale of those made from neodymium-iron-boron (NIB)." https://www.sciencealert.com/new-zealand-teen-loses-part-of-his-bowel-after-swallowing-nearly-200-magnets "A 13-year-old boy in New Zealand has had part of his bowel surgically removed after he ingested nearly 200 high-powered magnets. (2025)" OK, this teen was a fool, but it is not always a teen, it is not always...
Thread 'RIP Chen Ning Yang (1922-2025)'
https://en.wikipedia.org/wiki/Yang_Chen-Ning ( photo from http://insti.physics.sunysb.edu/~yang/ ) https://www.nytimes.com/2025/10/18/science/chen-ning-yang-dead.html https://www.bbc.com/news/articles/cdxrzzk02plo https://www.cpr.cuhk.edu.hk/en/press/mourning-professor-yang-chen-ning/ https://www.stonybrook.edu/commcms/physics/about/awards_and_prizes/_nobel_and_breakthrough_prizes/_profiles/yangc https://www.stonybrook.edu/commcms/physics/people/_profiles/yangc...

Similar threads

Back
Top