Peter wrote: The sidelengths of a polygon with $ 1994$ sides are $ a_{i} = \sqrt {i^2 + 4}$ $ \; (i = 1,2,\cdots,1994)$. Prove that its vertices are not all on lattice points. Proof (with a fixable flaw) We assume that $ i^2 + 2^2$ is a number which cannot be expressed as the sum of squares in more than one way (I have not checked this yet). We consider the $ S(x,y) = x + y \text{ (mod 2)}$ of a point. WLOG consider a point at the origin. Now we know that $ S(0,0) = 0$ We consider the movement between two points as $ \Delta (x,y) \leftarrow (x \pm 2,y \pm i) \text{ or } (x \pm i, y \pm 2)$. Considering this in terms of $ S(x,y)$, when $ 2|i$ then $ S(x,y) = S(\Delta ( x, y ) )$. We only need to consider the cases where $ i = 2k + 1$ Therefore we have 997 transforms which change $ S(x,y)$. We start with $ S(x,y) = 0, 1, 0, 1, 0, \cdots$ Since $ 2 \nmid 997$, $ S(\underbrace{\Delta \Delta \Delta \cdots \Delta}_{\text{1994 } \Delta \text{s}} ( 0, 0 ) ) = 1$ Since a polygon must connect back to the start, we have a contradiction. Fix Consider $ 2^2 + i^2 = k^2 + j^2 = 0 \text{ (mod 2)}$ Clearly $ i \equiv 0 \text{ (mod 2)}$ therefore $ S ( \Delta ( x, y ) ) = S ( x, y )$ Using $ T(x,y) = (x \pm k, y \pm j)$ we can see that the same thing happens. An equivalent argument holds for $ 1$ (mod $ 2$) Fix v2 Consider $ 2^2 + i^2 = k^2+j^2 = 0 \pmod{2}$ In this case, clearly $ i \equiv 0 \pmod{2}$. Therefore, $ S(\Delta(x,y)=S(x,y)$. If we consider the "modified" delta transform, $ \Delta_2 (x,y) \leftarrow (x \pm k, y \pm j ) \text{ or } (x \pm j, y \pm k)$ Either $ j^2 = k^2 = 0 \pmod{2}$ or $ j^2=k^2=1 \pmod{2}$. Since they are the same, $ S(\Delta(x,y)) = S(\Delta_2(x,y)) = S(x,y)$ Else consider $ 2^2 + i^2 = k^2+j^2 = 1 \pmod{2}$ In this case clearly $ i \equiv 1 \pmod{2}$. Therefore $ S( \Delta (x,y) ) = ( S ( x, y ) + 1 ) \pmod{2}$ However, either $ k^2 = 1, j^2 = 0$ or $ k^2 = 0, j^2 = 1$. WLOG consider the former. Using the same modified delta, we can see that $ S( \Delta_2 ( x, y ) ) = S ( x, y ) + 1$ which is the same as the usual delta transform (mod 2)

SimonM wrote: We assume that $ i^2 + 2^2$ is a number which cannot be expressed as the sum of squares in more than one way (I have not checked this yet). I'm sorry Simon, but $ 9^2 + 2^2 = 7^2 + 6^2$... By the way, \pmod{2} is better LaTeX than \text{ (mod 2)}.

Peter wrote: SimonM wrote: We assume that $ i^2 + 2^2$ is a number which cannot be expressed as the sum of squares in more than one way (I have not checked this yet). I'm sorry Simon, but $ 9^2 + 2^2 = 7^2 + 6^2$... By the way, \pmod{2} is better LaTeX than \text{ (mod 2)} Ah, I meant if $ i^2+2^2 \equiv 0 \pmod{2}$.

Sorry, I'm afraid this is not true: $ 26^2+2^2=14^2+22^2$...

Peter wrote: Sorry, I'm afraid this is not true: $ 26^2 + 2^2 = 14^2 + 22^2$... I just edited that bit of the proof, but the point is that the transform is still equivalent, no matter what the equivalent values are.

Now it is correct. Let me reword it a bit shorter though. Not bringing any new ideas, this is completely Simon's solution. Quote: Assume the contrary and denote $ S(x,y)=x+y\mod2$, and $ (x_0,y_0),(x_1,y_1),...,(x_{1994},x_{1994})$ the vertices of the polygon, with $ (x_0,y_0)=(x_{1994},y_{1994})$. Since $ z^2\equiv z\pmod{2}$ for all integer $ z$, we have \[ {i\equiv i+2\equiv i^2+2^2=a_i^2=(x_{i+1}-x_{i})^2+(y_{i+1}-y_{i})^2\equiv x_{i+1}+y_{i+1}-x_{i}-y_{i}=S(x_{i+1},y_{i+1})-S(x_i,y_i)\pmod{2}}.\] Now, since $ (x_0,y_0)=(x_{1994},y_{1994})$ we have $ \sum_{i=0}^{1993}S(x_{i+1},y_{i+1})=\sum_{i=0}^{1993}S(x_i,y_i)$, so the above gives us \[ 0=\sum_{i=0}^{1993}S(x_{i+1},y_{i+1})-S(x_i,y_i)\equiv\sum_{i=0}^{1993}i\equiv \sum_{i=0}^{1993}i\mod 2 = 997\pmod 2,\] contradiction.