Poor Mr. James didn't get to shake hands with everybody!

Suppose there are $n$ people in the faction opposed to Mr. James and $m$ people in the faction who are tolerant of him. Then the number of handshakes in the first group in $\dbinom{n}{2} = \dfrac{n(n - 1)}{2}$ , and in the second group, since there will be $m + 1$ people after Mr. James arrives, there will be $\dbinom{m + 1}{2} = \dfrac{(m + 1)m}{2}$ handshakes. So as there are $100$ handshakes in total, we need to find positive integers $m,n$ such that

$\dfrac{n(n - 1)}{2} + \dfrac{m(m + 1)}{2} = 100 \Longrightarrow n^{2} - n + m^{2} + m = 200 \Longrightarrow$

$\left(n - \dfrac{1}{2}\right)^{2} + \left(m + \dfrac{1}{2}\right)^{2} = 200 + \dfrac{1}{2} = \dfrac{401}{2} \Longrightarrow (2n - 1)^{2} + (2m + 1)^{2} = 802$ .

So we need to find two odd perfect squares which add to $802$ . Now as the largest odd perfect square less than $802$ is $27^{2} = 729$ we know that one of the squares will exceed $\dfrac{800}{2} = 400 = 20^{2}$ , so the larger of the potential perfect squares in our sum would have to be one of $21^{2} = 441, 23^{2} = 529, 25^{2} = 625$ or $27^{2} = 729$ . But the only one of these that yields another perfect square when subtracted from $802$ is $802 - 441 = 361 = 19^{2}$ , (the other differences being $273, 177$ and $73$ ).

So the only pair of odd perfect squares which add to $802$ are $19^{2}$ and $21^{2}$ . (A short cut would have been to note that $(20 - 1)^{2} + (20 + 1)^{2} = 2*400 + 2*1 = 802$ , but then we would still have to check for any other pairs.) Since we can assign these squares to either of $(2n - 1)^{2}$ or $(2m + 1)^{2}$ , we see that either $2n - 1 = 19, 2m + 1 = 21 \Longrightarrow n = 10, m = 10$ or $2n - 1 = 21, 2m + 1 = 19 \Longrightarrow n = 11, m = 9$ , and so the correct answer is $\boxed{\text{either 10 or 11}}$ .

Bonus question: In both cases there were $10 + 10 = 11 + 9 = 20$ people initially present.

• Since the total number of handshakes is 100, we compute $n$ such that 0< $\binom{n}{2}$ <100. The possible values of $n$ are $n=2,3,...,14$ . Now if i-1 is the number of people tolerant to Mr. james and j is the number people opposed to Mr. James, then $\binom{i}{2}$ + $\binom{j}{2}$ =100. We can easily check that the possible values of i and j are only 10 and 11.Thus, there are 20 people that were initially present in the meeting.