Lemma 1 There exist $ p-1$ positive integers not exceeding $ 2p^{2}$ whose pairwise sums are different. Proof We will take the integers $ a_{1},a_{2},...,a_{p-1}$ where for $ i = 1,2,3,...,p-1$, $ a_{i}= 2pi+\{i^{2}\}$ where by $ \{k\}$ we denote the residue of $ k$ $ \pmod p$. Assume we can find two distinct pairs of integers with equal pairwise sums, ie $ a_{x}+a_{y}= a_{z}+a_{t}$. Then: $ 2px+\{x^{2}\}+2py+\{y^{2}\}= 2pz+\{z^{2}\}+2pt+\{t^{2}\}$ (*) Because $ 2px+2py\leq a_{x}+a_{y}< 2p(x+y+1)$ as $ 0\leq\{x^{2}\}+\{y^{2}\}< 2p$, then $ a_{x}+a_{y}= a_{z}+a_{t}\Rightarrow x+y = z+t$, so $ x+y\equiv z+t\pmod p$ Now, substituting $ x+y=z+t$ into (*) we get $ \{x^{2}\}+\{y^{2}\}=\{z^{2}\}+\{t^{2}\}$. But then, $ x^{2}+y^{2}\equiv z^{2}+t^{2}\pmod p$, and so: $ 2(x^{2}+y^{2})-(x+y)^{2}\equiv 2(z^{2}+t^{2})-(z+t)^{2}\pmod p$ so: $ (x-y)^{2}\equiv (z-t)^{2}\pmod p$. Then, $ x-y\equiv z-t\pmod p$ or $ x-y = t-z\pmod p$, and if we use this and $ x+y\equiv z+t\pmod p$ we get $ (x,y) = (z,t)$ or $ (x,y) = (t,z)$ because $ x,y,z,t\in\{1,2,...,p-1\}$. But this contradicts the assumption that the two pairs of integers are distinct, hence there do not exist two distinct pairs with equal pairwise sums, and lemma 1 follows. Applying lemma 1 for $ p = 101$ and because $ 2*(101)^{2}= 20402 < 25000$ we get the result.