A Particular Square Number
Find a square number such that, when five is added or subtracted, the result is again a square number!
Labels: mathschallenge
posted by Zaux at
5:37 PM
Thursday, December 31, 2009
Links
Search
Share
- Tom Chrome Extension
- ToM @ Twitter
- MeeT ToM
- Daily Tricks Widget
- Vista Gadget 1.5
- iPhone Widget
- Windows XP Widget
- Google Toolbar Button
- Suggestions
Favorite Links
- Vista Sidebar Gadgets
- Widgets for Web 2.0
- C Sharp Tricks
- Interview Information
- Illustrator - Elar Alexander
Bookmark Us
Previous Posts
- Rabbits and More Rabbits
- Fashion Show
- The Mutilated Chessboard
- Strange Signs
- Mad Nomads
- Alien Ages
- Alphabet l
- Inscribed Polygons
- Hexagon In-Out
- Fair's Fair
19 Comments:
Not sure if this counts but
4 is a square number and
4+5=9=3^2
4-5=-1=i^2
Cam
Hi anonymous ... that's a neat clever solution ... but the intended solution involves only rational numbers.
Ok ... "i" is an imaginary number ... right? ... there is a solution not involving imaginary numbers. I certainly wouldn't pretend to argue mathematics with you ...
Cam ... Happy New Year!
(a/b)^2 - 5 = (c/d)^2 + 5
So, (ad)^2 = b^2 * (10 + c^2)
Then, ad/b = sqrt(10+c^2)
This implies that sqrt(10+c^2) is rational?
Have I missed something or made a false assumption?
I'll be interested to see the answer!
I can only come up with Cams solution also.
If you start the sequence of square numbers - 4-9-16-25-36-49-64 etc, there is a difference of 5 is between the 4 and the 9.
The answer is therefore 4 or less
If 4 then 4-5=-1 and 4+5=9 both fulfilling square-ness.
Happy New Year. I'm so glad to be back at work....
There is another solution ...
Zaux,
Happy New Years to you too.
For the record, only the 1st post was mine. It wasn't me arguing with you.
I only know applied maths. As such, I am blissfully ignorant of most number theory. I looked up "square numbers" on wikipedia after posting, and it confirms that square numbers indeed can not be negative i.e. their square rootmust be rational
I have a hunch as to what the angle on this problem may be... The search continues.
Cam
Here's a trick solution
x^5
x^(5-5)=X^0=1
sqrt(1)=1
X^(5+5)=X^10
sqrt(X^10)=X^5
5 being added or subtracted from the exponent rather than the number itself
Cam
I think the answer may be a square number with the number 5 added/taken away from it, ie 1521 (square of 39) and, with a 5 "subtracted", 121 (square of 11).
I am trying to come up with some sort of formula to cme up with asolution for both adding and subtracting a 5, I am going for a database solution...
Anyway, I'm going home, and will consider this further in the car!
Nice trick solution ... the addition and subtraction in the problem is a normal mathematical operation rather than a trick.
Karl,
As far as injecting 5 into a number to make another square number e.g. like A=121 to make B=1521
And then injecting another 5 to make another square number (no example found) C
B would be the number, where 5 was added or subtracted to make a square number
I already tried this and I could not find any solution for A under
1*10^15
Not to say, no solution exists by doing this, but it seems unlikely.
Cam
Consider fractions
Seems the activity has ceased on this problem ... try this number as the square:(a fraction)
1681/144 = (41/12)^2
You guys will understand some of the math better than I, but here goes: ( I hope that by shortening the explanation that I don't make a mistake)
find integer solutions of the equations:
x^2+5=y^2 and x^2-5=z^2
Using Fibonacci's congruous number form:
ab(a+b)(a-b) when a+b is even
and 4ab(a+b)(a-b)when a+b is odd
Fibonacci proved that congruous numbers are always divisible by 24.
He also showed that solutions of:
x^2+n=y^2 and x^2-n=z^2
can only be found if n is congruous.
In our problem n=5 and since 5 is not congruous, the problem is not solvable in integers.
However, a solution does exist with rational numbers:
from the facts that 720=12^2*5 is a congruous number (with a=5 and
b=4), and that 41^2=720=49^2 and 41^2-720=31^2, it follows by dividing both equations by 12^2 that:
x=41/12 y=49/12 z=31/12
thus x^2= (41/12)^2 = (1681/144)
Square number = 1681/144
Comment: I understand the math, except for the Fibonacci tie in to the problem ... oh well, hopefully it will make sense to some of you.
Sorry ... error:
The paragraph which begins "However, a solution does exist with rational numbers":
should read: 41^2+720=49^2
As you can see, the "=" sign should have been a "+" sign.
15
Meh...
I was following the definition in wikipedia "In mathematics, a square number, sometimes also called a perfect square, is an integer that can be written as the square of some other integer"
so I didn't think a^2/b^2 counted.
Cam
Cam ..I see your point ... apparently the puzzle source thought it to be ok.
Post a Comment
Links to this post:
Create a Link
<< Home