XbenX wrote: A positive integer is called hypotenuse if it can be represented as a sum of two squares of non-negative integers. Prove that any natural number greater than $10$ is the difference of two hypotenuse numbers. \[4a - 4 = a^2 - (a - 2)^2\]\[ 2a - 1 = a^2 - (a - 1)^2 \]\[ 4a - 3 = a^2 + 1^2 - (a - 1)^2 \]

GorgonMathDota wrote: XbenX wrote: A positive integer is called hypotenuse if it can be represented as a sum of two squares of non-negative integers. Prove that any natural number greater than $10$ is the difference of two hypotenuse numbers. \[4a - 4 = a^2 - (a - 2)^2\]\[ 2a - 1 = a^2 - (a - 1)^2 \]\[ 4a - 3 = a^2 + 1^2 - (a - 1)^2 \] $4a-3=a^2+1^2-(a-1)^2$?????

To be sure, what we want is more like \[2a-1=a^2-(a-1)^2\]and \[2a=(a^2+1^2)-(a-1)^2.\]We just need to make sure that all hypotuse numbers must be positive which is fine as soon as $a \ge 2$. But for $a=1$ we can certainly write $1=(1^2+1^2)-1^2$ and $2=2^2-(1^2+1^2)$. Does anyone understand why the problem only asks for numbers greater than $10$?

$$2x+1 = ((x+1)^2+0^2) - (x^2+0^2)$$$$2x= ((x+1)^2+0^2) - (x^2+1^2)$$

Dear XbenX, I am not sure to whom I should address my question. Nevertheless, in the "Contest Collections" page this problem is said to be proposed on Day 1. Whence the strange idea? St.Petersburg Olimpiad for grades 9-11 is a one-day event.