Chủ đề này có 1 trả lời

[Lao Hac]

Tìm các số $a,b$ nguyên tố thỏa mãn

$a(a+1) = b(b^2-1)$

[TranHungDao]

+) Xét $a=b=>...$

+) Xét $a\neq b=>(a,b)=1$

$a(a+1)=b(b^{2}-1)(*)$

From $(*)=>a(a+1)\vdots b=>a+1\vdots b$

$a+1=b.a_{1}(a_{1}\epsilon \mathbb{Z}^{+})$

$=>a.a_{1}=b^{2}-1=>b^{2}-1\vdots a=>(b-1)(b+1)\vdots a=>\begin{bmatrix}b-1\vdots a \\ b+1\vdots a \end{bmatrix}$

- T/H1:$\left\{\begin{matrix}a+1\vdots b \\ b+1\vdots a \end{matrix}\right.$

$=>(a+1)(b+1)\vdots ab=>a+b+1\vdots ab$

Since $a,b\epsilon \mathbb{P}=>a+b+1\geq ab<=>0< (a-1)(b-1)\leq 2=>\begin{bmatrix}(a-1)(b-1)=1 \\ (a-1)(b-1)=2 \end{bmatrix} ...$

- T/H2: $\left\{\begin{matrix}a+1\vdots b \\ b-1\vdots a \end{matrix}\right.$

$b-1=a.b_{1}(b_{1}\epsilon \mathbb{Z}^{+})=>b=ab_{1}+1$

$=>a+1\vdots ab_{1}+1=>ab_{1}+b_{1}\vdots ab_{1}+1=>b_{1}-1\vdots ab_{1}+1$

Since $b_{1}\epsilon \mathbb{Z}^{+}=>b_{1}-1\geq 0$

+ $b_{1}-1=0...$

+ $b_{1}-1> 0=>b_{1}-1\geq ab_{1}+1=>b_{1}(1-a)\geq 2$ hoang đường do $a> 1,b_{_{1}}\geq 1$