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, 08:48 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, 08:58 AMtopsquark
- Oct 4th 2007, 09:06 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

1) So must be divisible by c. Similarly from the second equation 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, 09:13 AMearboth
- Oct 4th 2007, 09:15 AMtopsquark
- Oct 4th 2007, 09:30 AMtopsquark
So again we may divide each of these by 2:

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

It looks to me like:

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, 11:19 AMHelloworld
- Oct 5th 2007, 08:32 AMHelloworld
ok, now I know how can I do that ;)Topic should be closed ;)

- Oct 5th 2007, 02:51 PMJhevon