Consider the quadratic equation \[ x^2 - 5x + k = 0. \] For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \[ D = b^2 - 4ac. \] Here, \(a = 1\), \(b = -5\), \(c = k\), so \[ D = 25 - 4k. \] Roots are \[ x = \frac{5 \pm \sqrt{25 - 4k}}{2}. \] For the roots to be integers:
1. The discriminant \(25 - 4k\) must be a non-negative perfect square.
2. The numerator \(5 \pm \sqrt{25 - 4k}\) must be even (so that division by 2 gives an integer).
Step 1: Range of $k$ from non-negative discriminant. We need \[ 25 - 4k \ge 0 \Rightarrow 4k \le 25 \Rightarrow k \le 6.25. \] Since \(k\) is a non-negative integer, \[ k \in \{0,1,2,3,4,5,6\}. \]
Step 2: Check for perfect square discriminants. Compute \(D = 25 - 4k\) for each possible \(k\): k = 0 & D = 25 \((=5^2)\)
k = 1 & D = 21 (not a perfect square)
k = 2 & D = 17 (not a perfect square)
k = 3 & D = 13 (not a perfect square)
k = 4 & D = 9 \((=3^2)\)
k = 5 & D = 5 (not a perfect square)
k = 6 & D = 1 \((=1^2)\)
Thus, \(D\) is a perfect square for \[ k = 0,\ 4,\ 6. \]
Step 3: Verify roots are integers in these cases. - For \(k = 0\): \[ x^2 - 5x = 0 \Rightarrow x(x - 5) = 0 \Rightarrow x = 0, 5\quad (\text{integers}). \] - For \(k = 4\): \[ x^2 - 5x + 4 = 0 \Rightarrow x = \frac{5 \pm \sqrt{9}}{2} = \frac{5 \pm 3}{2} \Rightarrow x = 4, 1\quad (\text{integers}). \] - For \(k = 6\): \[ x^2 - 5x + 6 = 0 \Rightarrow x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2} \Rightarrow x = 3, 2\quad (\text{integers}). \] All three values of \(k\) give integer roots. Therefore, the number of non-negative integer values of \(k\) for which the quadratic has only integer roots is \[ \boxed{3}. \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: