ubavic je napisao:[inlmath]7.[/inlmath] Odrediti sve cele brojeve [inlmath]a[/inlmath] za koje su izrazi [inlmath]96 + a[/inlmath] i [inlmath]5 + a[/inlmath] kubovi celih brojeva.
[dispmath]\begin{array}{ll}
96+a=x^3 & \left(1\right) \\
5+a=y^3 & \left(2\right)
\end{array}[/dispmath][dispmath]\left(1\right)-\left(2\right)\quad\Rightarrow\quad x^3-y^3=91[/dispmath][dispmath]\left(x-y\right)\left(x^2+xy+y^2\right)=91[/dispmath][inlmath]x-y=k[/inlmath]
[inlmath]y=x-k[/inlmath][dispmath]k\left[x^2+x\left(x-k\right)+\left(x-k\right)^2\right]=91[/dispmath][dispmath]k\left(x^2+x^2-kx+x^2-2kx+k^2\right)=91[/dispmath][dispmath]k\left(3x^2-3kx+k^2\right)=91[/dispmath]Faktori broja [inlmath]91[/inlmath] su [inlmath]-91[/inlmath], [inlmath]-13[/inlmath], [inlmath]-7[/inlmath], [inlmath]-1[/inlmath], [inlmath]1[/inlmath], [inlmath]7[/inlmath], [inlmath]13[/inlmath] i [inlmath]91[/inlmath]. Prema tome, on se može napisati na sledeće načine:
[dispmath]\begin{array}{ll}
91\mathop=\left(-91\right)\cdot\left(-1\right) & \Rightarrow & k=-91, & 3x^2-3kx+k^2=-1 \\
91\mathop=\left(-13\right)\cdot\left(-7\right) & \Rightarrow & k=-13, & 3x^2-3kx+k^2=-7 \\
91\mathop=\left(-7\right)\cdot\left(-13\right) & \Rightarrow & k=-7, & 3x^2-3kx+k^2=-13 \\
91\mathop=\left(-1\right)\cdot\left(-91\right) & \Rightarrow & k=-1, & 3x^2-3kx+k^2=-91 \\
91\mathop=1\cdot 91 & \Rightarrow & k=1, & 3x^2-3kx+k^2=91 \\
91\mathop=7\cdot 13 & \Rightarrow & k=7, & 3x^2-3kx+k^2=13 \\
91\mathop=13\cdot 7 & \Rightarrow & k=13, & 3x^2-3kx+k^2=7 \\
91\mathop=91\cdot 1 & \Rightarrow & k=91, & 3x^2-3kx+k^2=1
\end{array}[/dispmath][dispmath]3x^2-3kx+k^2=\frac{91}{k}\quad /\cdot k[/dispmath][dispmath]3kx^2-3k^2x+k^3=91[/dispmath][dispmath]3kx^2-3k^2x+k^3-91=0[/dispmath]Pošto rešenje ove jednačine po [inlmath]x[/inlmath] mora biti ceo broj, diskriminanta ove jednačine mora biti potpun kvadrat:[dispmath]D=9k^4-4\cdot 3k\left(k^3-91\right)[/dispmath][dispmath]D=9k^4-4\cdot 3k^4+4\cdot 3k\cdot 91[/dispmath]equation]D=-3k^4+4\cdot 3k\cdot 91[/equation][dispmath]D=3k\left(-k^3+4\cdot 91\right)[/dispmath][dispmath]D=3k\left(364-k^3\right)[/dispmath][dispmath]D>0\quad\Rightarrow\quad k\left(364-k^3\right)>0\quad\Rightarrow\quad\left(k>0\;\land\;364-k^3>0\right)\;\lor\;\left(k<0\;\land\;364-k^3<0\right)[/dispmath][dispmath]\left(k>0\;\land\;k^3<364\right)\;\lor\;\left(k<0\;\land\;k^3>364\right)[/dispmath][dispmath]\left(k>0\;\land\;k\le 7\right)\;\lor\;\cancelto{\bot}{\left(k<0\;\land\;k\ge 8\right)}[/dispmath][dispmath]\Rightarrow\quad k=1\quad\lor\quad k=7[/dispmath][inlmath]\underline{k=1}[/inlmath]
[inlmath]D=3\cdot 363=1089=33^2[/inlmath]
potpun kvadrat
[inlmath]3kx^2-3k^2x+k^3-91=0[/inlmath]
[inlmath]3x^2-3x+1-91=0[/inlmath]
[inlmath]3x^2-3x-90=0[/inlmath]
[inlmath]x_{1,2}=\frac{3\pm 33}{6}[/inlmath]
[inlmath]x_1=-5\quad\Rightarrow\quad 96+a=\left(-5\right)^3=-125\quad\Rightarrow\quad a=-125-96\quad\Rightarrow\quad\underline{a=-221}[/inlmath]
[inlmath]x_2=6\quad\Rightarrow\quad 96+a=6^3=216\quad\Rightarrow\quad a=216-96\quad\Rightarrow\quad\underline{a=120}[/inlmath]
[inlmath]\underline{k=7}[/inlmath]
[inlmath]D=3\cdot 7\left(364-7^3\right)=21\cdot 21\mathop=21^2[/inlmath]
potpun kvadrat
[inlmath]3kx^2-3k^2x+k^3-91=0[/inlmath]
[inlmath]21x^2-147x+343-91=0[/inlmath]
[inlmath]21x^2-147x+252=0[/inlmath]
[inlmath]x_{1,2}=\frac{147\pm 21}{42}[/inlmath]
[inlmath]x_1=3\quad\Rightarrow\quad 96+a=3^3=27\quad\Rightarrow\quad a=27-96\quad\Rightarrow\quad\underline{a=-69}[/inlmath]
[inlmath]x_2=4\quad\Rightarrow\quad 96+a=4^3=64\quad\Rightarrow\quad a=64-96\quad\Rightarrow\quad\underline{a=-32}[/inlmath]
[dispmath]\enclose{box}{a\in\left\{-221,-69,-32,120\right\}}[/dispmath]