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.
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
on triangular doubles
