All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Double a triangle (Posted on 2015-02-19) Difficulty: 3 of 5
In the sequence of triangular numbers, some numbers are twice another.

For example t(20)=210 which is twice t(14)=105.

Characterize all such numbers.

Easy bonus: Explain why (except for the trivial case) there are no square numbers that are twice another.

No Solution Yet Submitted by Jer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution on triangular doubles | Comment 2 of 3 |
      

Lemma:

The equation  
   x^2=2y^2+1       (Pell eq.)

has an infinite number of integer solutions.                                           

Proof:  
  Clearly (1,0) is a solution.
It is obvious that if (x,y) is a solution so is (3x+4y, 2x+3y), therefore  (1,0)  generates a chain of valid solutions, each derived from the previous  one: 
  (1,0);  (3,2);  (17,12); (99,70);… etc

Back to our problem:

n/m=2*(m+1)/(n+1)    let us call  n/m=p/q

so n=pm/q   and n+1=2q*(m+1/p)

and m(p^2-2q^2) = 2q^2-pq

 

Evaluating  (p,q) by solving the Pel. Eq.

we get integer values  for m and n:

m=q*(2q-p)  and n= p*(2q-p) 

Example:

 (p,q)=(17,12) generates ( m,n)=(84,119)

 

Bonus  question, having nothing in common with the topic discussed above, can be resolved by an easy proof:

If m and n are integers (any inegers) then m/n is a ratoinal number, but m^2/n^2=2  forces
m=n*sqrt(2) which is irrational.
Proven by contradiction.

 

 

 


 

Edited on February 19, 2015, 3:31 pm
  Posted by Ady TZIDON on 2015-02-19 15:14:31

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information