+3점 · ( k + r 2 k ) 2 = k 2 + r + r 2 4 k 2 ≥ k 2 + r = n \displaystyle \left(k+\frac r{2k}\right)^2=k^2+r+\frac{r^2}{4k^2}\ge k^2+r=n ( k + 2 k r ) 2 = k 2 + r + 4 k 2 r 2 ≥ k 2 + r = n k + r 2 k ≥ n \displaystyle k+\frac r{2k}\ge\sqrt n k + 2 k r ≥ n +4점 · 또한 r = n − k 2 < ( k + 1 ) 2 − k 2 = 2 k + 1 \displaystyle r=n-k^2<(k+1)^2-k^2=2k+1 r = n − k 2 < ( k + 1 ) 2 − k 2 = 2 k + 1 r ≤ 2 k \displaystyle r\le2k r ≤ 2 k +3점 · ( k + r 2 k ) 2 ≤ k 2 + r + 1 = n + 1 \displaystyle \left(k+\frac r{2k}\right)^2\le k^2+r+1=n+1 ( k + 2 k r ) 2 ≤ k 2 + r + 1 = n + 1 k + r 2 k ≤ n + 1 \displaystyle k+\frac r{2k}\le\sqrt{n+1} k + 2 k r ≤ n + 1 (2-1) ( k + r 2 k ) 2 = k 2 + r + r 2 4 k 2 ≥ k 2 + r = n \displaystyle \left(k+\frac r{2k}\right)^2=k^2+r+\frac{r^2}{4k^2}\ge k^2+r=n ( k + 2 k r ) 2 = k 2 + r + 4 k 2 r 2 ≥ k 2 + r = n k + r 2 k ≥ n \displaystyle k+\frac r{2k}\ge\sqrt n k + 2 k r ≥ n r = n − k 2 < ( k + 1 ) 2 − k 2 = 2 k + 1 \displaystyle r=n-k^2<(k+1)^2-k^2=2k+1 r = n − k 2 < ( k + 1 ) 2 − k 2 = 2 k + 1 r ≤ 2 k \displaystyle r\le2k r ≤ 2 k 따라서 ( k + r 2 k ) 2 = k 2 + r + r 2 4 k 2 ≤ k 2 + r + 1 = n + 1 \displaystyle \left(k+\frac r{2k}\right)^2=k^2+r+\frac{r^2}{4k^2}\le k^2+r+1=n+1 ( k + 2 k r ) 2 = k 2 + r + 4 k 2 r 2 ≤ k 2 + r + 1 = n + 1 이고 k + r 2 k ≤ n + 1 \displaystyle k+\frac r{2k}\le\sqrt{n+1} k + 2 k r ≤ n + 1
+6점 · ( k + r + 1 2 ( k + 1 ) ) 2 − n = ( r − 2 k − 1 ) 2 4 ( k + 1 ) 2 ≥ 0 \displaystyle \left(k+\frac{r+1}{2(k+1)}\right)^2-n=\frac{(r-2k-1)^2}{4(k+1)^2}\ge0 ( k + 2 ( k + 1 ) r + 1 ) 2 − n = 4 ( k + 1 ) 2 ( r − 2 k − 1 ) 2 ≥ 0 k + r + 1 2 ( k + 1 ) ≥ n \displaystyle k+\frac{r+1}{2(k+1)}\ge\sqrt n k + 2 ( k + 1 ) r + 1 ≥ n (2-2) n = k 2 + r \displaystyle n=k^2+r n = k 2 + r ( k + r + 1 2 ( k + 1 ) ) 2 − n = ( k + r + 1 2 ( k + 1 ) ) 2 − k 2 − r = r + 1 2 ( k + 1 ) ( 2 k + r + 1 2 ( k + 1 ) ) − r = r 2 − 2 ( 2 k + 1 ) r + ( 4 k 2 + 4 k + 1 ) 4 ( k + 1 ) 2 = ( r − 2 k − 1 ) 2 4 ( k + 1 ) 2 ≥ 0 \displaystyle \begin{aligned}\left(k+\frac{r+1}{2(k+1)}\right)^2-n&=\left(k+\frac{r+1}{2(k+1)}\right)^2-k^2-r=\frac{r+1}{2(k+1)}\left(2k+\frac{r+1}{2(k+1)}\right)-r\\&=\frac{r^2-2(2k+1)r+(4k^2+4k+1)}{4(k+1)^2}=\frac{(r-2k-1)^2}{4(k+1)^2}\ge0\end{aligned} ( k + 2 ( k + 1 ) r + 1 ) 2 − n = ( k + 2 ( k + 1 ) r + 1 ) 2 − k 2 − r = 2 ( k + 1 ) r + 1 ( 2 k + 2 ( k + 1 ) r + 1 ) − r = 4 ( k + 1 ) 2 r 2 − 2 ( 2 k + 1 ) r + ( 4 k 2 + 4 k + 1 ) = 4 ( k + 1 ) 2 ( r − 2 k − 1 ) 2 ≥ 0 따라서 제시문 (나)에 의하여 k + r + 1 2 ( k + 1 ) ≥ n \displaystyle k+\frac{r+1}{2(k+1)}\ge\sqrt n k + 2 ( k + 1 ) r + 1 ≥ n
+4점 · ( k + r + 1 2 ( k + 1 ) ) 2 − n − 1 = ( r − 4 k − 3 ) ( r + 1 ) 4 ( k + 1 ) 2 ≤ 0 \displaystyle \left(k+\frac{r+1}{2(k+1)}\right)^2-n-1=\frac{(r-4k-3)(r+1)}{4(k+1)^2}\le0 ( k + 2 ( k + 1 ) r + 1 ) 2 − n − 1 = 4 ( k + 1 ) 2 ( r − 4 k − 3 ) ( r + 1 ) ≤ 0 k + r + 1 2 ( k + 1 ) ≤ n + 1 \displaystyle k+\frac{r+1}{2(k+1)}\le\sqrt{n+1} k + 2 ( k + 1 ) r + 1 ≤ n + 1 또한, (2-1)에서 r ≤ 2 k \displaystyle r\le2k r ≤ 2 k ( k + r + 1 2 ( k + 1 ) ) 2 − n − 1 = ( r − 2 k − 1 ) 2 4 ( k + 1 ) 2 − 1 = ( r − 4 k − 3 ) ( r + 1 ) 4 ( k + 1 ) 2 ≤ 0 \displaystyle \left(k+\frac{r+1}{2(k+1)}\right)^2-n-1=\frac{(r-2k-1)^2}{4(k+1)^2}-1=\frac{(r-4k-3)(r+1)}{4(k+1)^2}\le0 ( k + 2 ( k + 1 ) r + 1 ) 2 − n − 1 = 4 ( k + 1 ) 2 ( r − 2 k − 1 ) 2 − 1 = 4 ( k + 1 ) 2 ( r − 4 k − 3 ) ( r + 1 ) ≤ 0 이다. 따라서 제시문 (나)에 의하여 k + r + 1 2 ( k + 1 ) ≤ n + 1 \displaystyle k+\frac{r+1}{2(k+1)}\le\sqrt{n+1} k + 2 ( k + 1 ) r + 1 ≤ n + 1
+10점 · 짝수와 홀수로 나누어 생각하면 2점, 올바른 p , q \displaystyle p,q p , q Correction. r \displaystyle r r p \displaystyle p p q \displaystyle q q
(2-3) 주어진 자연수 n \displaystyle n n r \displaystyle r r n ≤ k 2 + r 2 k ≤ n + 1 \displaystyle \sqrt n\le\frac{k^2+\frac r2}{k}\le\sqrt{n+1} n ≤ k k 2 + 2 r ≤ n + 1 이므로 p = k 2 + r 2 \displaystyle p=k^2+\frac r2 p = k 2 + 2 r q = k \displaystyle q=k q = k 주어진 자연수 n \displaystyle n n r \displaystyle r r n ≤ k ( k + 1 ) + r + 1 2 k + 1 ≤ n + 1 \displaystyle \sqrt n\le\frac{k(k+1)+\frac{r+1}2}{k+1}\le\sqrt{n+1} n ≤ k + 1 k ( k + 1 ) + 2 r + 1 ≤ n + 1 이므로 p = k ( k + 1 ) + r + 1 2 \displaystyle p=k(k+1)+\frac{r+1}2 p = k ( k + 1 ) + 2 r + 1 q = k + 1 \displaystyle q=k+1 q = k + 1
수리논술, 배움에서 논증의 완성까지.