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In a related question, I defined $T_{n,k}$ to be the remainder of dividing triangle number $T_n$ by $k$. That question was too easily resolved, by counterexample, so let me loosen the condition. Is the following conjecture true?

Conjecture: There exists some natural number $L > 3$, such that, for all values of $k >L, i < k$, there exists some $n$ such that $T_{n, k} = i$.

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    It was already shown in an answer to your previous question that whenever $k$ is a prime greater than three, $T_{n,k}=i$ is not solvable for all $i$L$ you choose, there will _always_ be a prime that is greater. So I'm not sure why you even asked this question. – 2017-02-21
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    it doesn't work for $3$ either. – 2017-02-21

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No, it is false, if $p$ is an odd prime then we have:

$n(n+1)/2\equiv x\bmod p \iff n(n+1)\equiv 2x\iff 4n^2+4n\equiv 8x\iff (2n+1)^2\equiv 8x+1\bmod p$.

So we need for $8x+1$ to be a quadratic residue $\bmod p$. The function $f:\mathbb Z_p \rightarrow \mathbb Z_p$ given by $f(x)=8x+1$ is clearly bijective so we can always find $x$ such that $8x+1$ is not a quadratic residue $\bmod p$.

So it is false for every odd prime.

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    I take it the bijectivity of that function stems from the fact that $8$ is coprime to any odd prime? – 2017-02-23
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    Yes thats exactly It. – 2017-02-23