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As in the title. I've read that 2016 cannot be expressed in such form, but I've completely no idea, how could this fact be proven.

3 Answers3

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HINT: $2016$ is itself a triangular number since $$2016 = \frac{63.64}{2}$$ Is it possible that the difference between two triangular numbers is prime?

Crostul
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    To make the OP appreciate a certain subtlety here let me point out that $$\frac{64\cdot65}2-\frac{62\cdot63}2=127$$ is a prime. +1 of course – Jyrki Lahtonen Dec 23 '15 at 11:03
  • @JyrkiLahtonen I usually do not make personal comments, but I must say that you are such a blessing to the community. +1 to your comment and the answer. – Shailesh Dec 23 '15 at 11:22
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Assume $2016 = p + t$, where $t$ is triangular, so $t$ can be written as $k(k + 1)/2$. So $$ \begin{align} 2016 &= p + \frac{k(k+1)}{2} \\ &= \frac{63 \cdot 64}{2} = \frac{63 \cdot (63 + 1)}{2} \\ \implies p &= \frac{63 \cdot (63 + 1)}{2} - \frac{k(k+1)}{2} \\ &= \frac{(63-k)(64+k)}{2} \\ &= (63-k)(32+\frac{k}{2}) \end{align} $$

If $k$ is even ($k = 2e$), both $63 - k$ and $32 + e$ are factors of $p$. Thus in order to make $p$ a prime number, $k$ can only be 62, but then $32 + e = 32 + 62/2 = 63 = 7 \cdot 9$, so $k$ cannot be even.

If $k$ is odd number($k = 2o + 1$), then $p = (62-2o)(32 + o + 1/2) = (31-o)(65+2o)$. Similarly, to make $p$ a prime number, $o$ can only be 30, but $65 + 2*30 = 125 = 5^3$. So we proved that no such $k$ exists, and hence no such $p$ exists.

shardulc
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wenlu sun
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    A few other possibilities need to be checked. If $63-k=-1$ then $k=64$ so $p=-64$, which doesn't work. If $32+\frac{k}{2}=\pm 1$ then $k=-62$ or $k=-66$, so $p=125$ or $p=-129$, which doesn't work. If $31-o=-1$ then $o=32$ so $p=-129$, which doesn't work. If $65+2o=\pm 1$ then $o=-33$ or $o=-32$, so $p=-64$ or $p=63$, which doesn't work. – J.G. Dec 23 '15 at 12:51
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Just an observation, it seems that what @Crostul: and @ Jyrki Lahtonen: are saying is that a prime is a difference of two triangular numbers in only two ways:

\begin{eqnarray} p &= &(1+2+\cdot + p)- (1+2+\cdots + (p-1)) \\ p &= & 2 n-1= (1+2+\cdots + n) - (1+ 2 + \cdots + n-2) \end{eqnarray}

so a sum of a string consecutive positive integers of length $1$ or $2$.

(thanks to @ Jyrki Lahtonen: for pointing out the first variant)

orangeskid
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  • Yup. The difference of two triangular numbers is a sum of $m$ consecutive integers. If $m$ is divisible by an odd prime $p$, then so is the said sum. If $m$ is divisible by four, then the said sum is even. Ergo, if that sum is a prime we must have $m=1$ or $m=2$. There may be a simpler way of thinking about it. – Jyrki Lahtonen Dec 23 '15 at 19:44
  • @Jyrki Lahtonen: Oh yes, one factor can be $1$ instead of $2$. Correct, the string of consecutive integers has length $1$ or $2$. Thank you! – orangeskid Dec 23 '15 at 20:01