I found that the series represented by(adsbygoogle = window.adsbygoogle || []).push({}); Anever includes a perfect square if and only if the prime factorization of_{n}= n∗(n+1)/2 - k8∗k+1includes a prime factor,p, to theith power wherep = +/−3 mod 8andiisoddand that this can be proved modp^{(i + 1)}

For instance, [tex]8* 4 + 1 = 3^1 * 11^1[/tex] so we can provemod 9thatn*(n+1)/2 - 4is never a square. Thesquare residues mod 9 are {0,1,4,7}but forn = 0 to 8the residues ofn*(n+1)/2 - 4 mod 9are{5,6,8,2,6,2,8,6,5}which are not any of the square residues. Thusn*(n+1)/2 - 4can never be a square for integern.

I also found that if, to the8∗k+1includes a prime factor, pith power wherep = +/−3 mod 8andi is oddthatk * (2*m+1)is a^{2}+m(m+1)/2k'such that8k' +1also containspto an odd power in the prime factorization. I was wondering if there is any higher algebra that could prove my findings.

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# When n*(n+1)/2 - k is never a square

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