Question by Dao Trong Luan

Dao Trong Luan Coordinator

03/12/2017 at 10:06

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Find 3 positive integer number a,b,c , know:

$\left\{{}\begin{matrix}a^3+3a^2+5=5^b\\a+3=5^c\end{matrix}\right.$

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Phan Thanh Tinh Coordinator 03/12/2017 at 17:31

$a^3+3a^2+5=5^b\Leftrightarrow a^2\left(a+3\right)+5=5^b\Leftrightarrow a^2.5^c+5=5^b$

$a\ge1\Rightarrow a^3+3a^2+5\ge1^3+3.1^2+5=9\Rightarrow5^b\ge9\Rightarrow b\ge2$

Moreover : $a+3\ge4\Rightarrow5^c\ge4\Rightarrow c\ge1$

So, we can divide both sides by 5 : $a^2.5^{c-1}+1=5^{b-1}$

Since $b-1\ge1;c-1\ge0$, in modulo 10, we have :

$5^{b-1}\equiv5\Rightarrow a^2.5^{c-1}\equiv4$

If $c-1\ge1$ : $5^{c-1}\equiv5\Rightarrow a^2.5^{c-1}\equiv0$ or $a^2.5^{c-1}\equiv5$ (unsatisfied)

$\Rightarrow c-1=0\Rightarrow c=1\Rightarrow a+3=5\Rightarrow a=2\Rightarrow5^b=25\Rightarrow b=2$

Hence, a = 2 ; b = 2 ; c = 1
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FA Liên Quân Garena 30/12/2017 at 21:46

$a^{3} + 3 a^{2} + 5 = 5^{b} \Leftrightarrow a^{2} (a + 3) + 5 = 5^{b} \Leftrightarrow a^{2} {.5}^{c} + 5 = 5^{b}$

$a \geq 1 \Rightarrow a^{3} + 3 a^{2} + 5 \geq 1^{3} + {3.1}^{2} + 5 = 9 \Rightarrow 5^{b} \geq 9 \Rightarrow b \geq 2$

Moreover : $a + 3 \geq 4 \Rightarrow 5^{c} \geq 4 \Rightarrow c \geq 1$

So, we can divide both sides by 5 : $a^{2} {.5}^{c - 1} + 1 = 5^{b - 1}$

Since $b - 1 \geq 1; c - 1 \geq 0$

, in modulo 10, we have :

$5^{b - 1} \equiv 5 \Rightarrow a^{2} {.5}^{c - 1} \equiv 4$

If $c - 1 \geq 1$

: $5^{c - 1} \equiv 5 \Rightarrow a^{2} {.5}^{c - 1} \equiv 0$ or $a^{2} {.5}^{c - 1} \equiv 5$

(unsatisfied)

$\Rightarrow c - 1 = 0 \Rightarrow c = 1 \Rightarrow a + 3 = 5 \Rightarrow a = 2 \Rightarrow 5^{b} = 25 \Rightarrow b = 2$

Hence, a = 2 ; b = 2 ; c = 1

Dao Trong Luan Coordinator

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