Hi I'm glad to be a part of Physics forum. I am retired.

[OlderOwl]

I retired from the US Patent and Trademark Office after serving 35+ years there. I had a passion for Math in elementary and high school but never had much more than the minimum required calculus and differential equations etc. required for engineering, having taken nighttime instruction at George Washington University while working as a full time aide to patent examiners. I became a Primary US patent Examiner myself after meeting minimum education and job skills requirements. My current passion is triangular numbers and elementary number theory. I found that the series represented by $$A_n = n*(n+1)/2 + k$$ always includes a perfect square if and only if the prime factorization of $$8*k +1$$ does not include a prime factor to the ith power where $$p^i = +/- \ 3 \ mod\ 8$$. For instance, if $$k = 1 ; 8*k +1 = 3^2$$ which is not equal to 3 or 5 mod 8. Thus there are perfect squares that are a triangular number plus 1. However, 8*4+1 = 33 which includes 3 and 11 each to an odd power, thus n*(n+2)/2 + 4 can never be a perfect square. This can be proven by using $$p^{(i+1)}$$ as the modulus to show that n*(n+1)/2 + k is never a square residue where $$p^i = +/-\ 3 \mod\ 8$$ is a part of the prime factorization of 8k +1.

 

[Greg Bernhardt]

Welcome to PF!

 

[OlderOwl]

My big typo,. I should have typed [Tex]n*(n-1)/2 - k [/Tex] where I typed n*(n+1)/2+k for instance 0 is a triangular number while 0 + 4 is a square 0 - 4 is not a square. Neither is n*(n+n)/2 - 4 a square for any integer n. Proof since 3^1 is part of the prime factorization of 8*4 + 1 and equals +/- 3 mod 8 we use mod 3^2 (mod 9): n*(n+1)/2 - 4≡{5,6,8,2,6,2,8,6,5} mod 9 for n ={0,1,2,3,4,5,6,7,8} but the square residues mod 9 are {0,1,4,7}. If the moderators of this forum can change my original post to reflect this correction or show me how to, I would appreciate it.

 

[Greg Bernhardt]

Please post your question in the correct forum :)