Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance ;)

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- Oct 4th 2007, 07:48 AMHelloworldSimultaneous equations
Hello, I've a problem with equations. It's very hard to me. Could u give the solutions with some steps?10(b^3)=8c+2(a^5)10(c^3)=8a+2(b^5)10(a^3)=8b +2(c^5)I think, that:x,y,z=0x,y,z=1but what's next?x=? y=? z=?I need help.Thanks in advance ;)

- Oct 4th 2007, 07:58 AMtopsquark
Let's write this out a little more intelligibly:

$\displaystyle 10b^3=8c+20a^5c^3=8a+20a^3b^5=8b+2c^5$

I suppose as a first step I'd write out:

$\displaystyle 10b^3 = 8c + 20a^5c^3 \implies 5b^3 = 4c + 10a^5c^3$

$\displaystyle 10b^3 = 8a + 20a^3b^5 \implies 5b^3 = 4a + 10a^3b^5$

$\displaystyle 10b^3 = 8b + 2c^5 \implies 5b^3 = 4b + c^5$

-Dan - Oct 4th 2007, 08:06 AMtopsquark
I presume a, b, and c are to be positive integers.

I can't finish this for you, but I will note a couple of points that you might be able to put together.

The first equation says that

$\displaystyle 5b^3 = c(4 + 10a^5c^2)$

1) So $\displaystyle 5b^3$ must be divisible by c. Similarly from the second equation $\displaystyle 5b^3$ must be divisible by a.

2) The RHS of both the first and second equations is divisible by 2. Thus b is divisible by 2.

3) Using 2), we see from the third equation that c is also divisible by 2.

(So note that your solution of a = 1, b = 1, and c = 1 must be incorrect.)

See if these observations get you anywhere.

-Dan - Oct 4th 2007, 08:13 AMearboth
- Oct 4th 2007, 08:15 AMtopsquark
- Oct 4th 2007, 08:30 AMtopsquark
So again we may divide each of these by 2:

$\displaystyle \begin{array}{l}5b^3=4c+a^5 \\ 5c^3=4a+b^5 \\ 5a^3=4b+c^5 \end{array}$

I guess this would explain your (now valid) a = 1, b = 1, c = 1 solution.

It looks to me like:

$\displaystyle 5b^3=4c+a^5 \implies 4c = 5b^3 - a^5$

implies that a and b are either both even or both odd.

Similarly the second equation implies that b and c are either both even or both odd.

Similarly the third equation, etc.

So it looks like either all of a, b, and c are even or all of a, b, and c are odd.

-Dan - Oct 4th 2007, 10:19 AMHelloworld
- Oct 5th 2007, 07:32 AMHelloworld
ok, now I know how can I do that ;)Topic should be closed ;)

- Oct 5th 2007, 01:51 PMJhevon