Question: Find all integers x, y, z such that > x^2 + y^2 = -z^2. > Why it’s impossible: For any integer, a square is always nonnegative, so: * x^2 \ge 0 * y^2 \ge 0 * z^2 \ge 0 The left side, x^2 + y^2, is therefore at least 0, while the right side, -z^2, is at most 0. They can only be equal if both sides are 0, which happens only when: x = y = z = 0. So if the problem instead asks for nonzero integers, it has no solution. If you want something that looks solvable but is actually impossible (an olympiad-style trick question), try this: Challenge: Find positive integers a, b, c satisfying > a^2 + b^2 + c^2 = -1. > This is impossible because the left side is always nonnegative, so it can never equal -1.

Let f:\mathbb{R}\rightarrow\mathbb{R} satisfy all of the following simultaneously. Part I For every real number x, f(x+1)=f(x)+2x+1 and f(x+y)+f(x-y)=2f(x)+2f(y) for every real x,y. Also, \int_0^1 f(x)\,dx=0 and \sum_{n=1}^{\infty}\frac{f(n)}{2^n}=100. ⸻ Part II Let A= \begin{pmatrix} f(1)&1&1\\ 1&f(2)&1\\ 1&1&f(3) \end{pmatrix}. Find every eigenvalue of A, prove whether A is positive definite, compute A^{100}, and determine every integer k such that \det(A^k)=2026. ⸻ Part III Define g(x)= \int_0^x \frac{\sin(f(t))}{1+t^2}\,dt. Prove or disprove that g(x)=g(-x) for every real number. Then determine every point where g''(x)=0. ⸻ Part IV Suppose p is a prime satisfying p^2+2=f(p). Prove whether infinitely many such primes exist. If only finitely many exist, determine the largest one. ⸻ Part V Let S=\sum_{n=1}^{\infty} \frac{(-1)^n}{n!} \left( \int_0^n f(x)\,dx \right). Evaluate S exactly. ⸻ Part VI A graph G has * 100 vertices, * every vertex has