if a^3 + a = b^3 + b then a = b

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IProto

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I have to prove the following

Prove:
if a^3 + a = b^3 + b then a = b

I have tried setting the assumption all to one side and factoring it out and substituting in place of variables and all that fun stuff. Nothing seems to work.

Oh, a and b are both integers.

I don't need a full solution for this problem, I'm just looking for a little direction.
 
IProto said:
Prove, for integers a and b, that, if a^3 + a = b^3 + b then a = b
Click to expand...
This might help:

. . . . .a<sup>3</sup> + a = b<sup>3</sup> + b

. . . . .a<sup>3</sup> - b<sup>3</sup> = b - a

. . . . .(a - b)(a<sup>2</sup> + ab + b<sup>2</sup>) = -(a - b)

If a does not equal b, then a - b is non-zero, and you can divide it out. What does this leave you with?

IProto said:
I have tried setting the assumption all to one side and factoring it out and substituting in place of variables and all that fun stuff. Nothing seems to work.
Click to expand...
In the future, kindly include what you've tried, so the tutors can see what you're doing. Often enough, you'll learn that you were on the right track! :wink:

Thank you! :D

Eliz.
 
Hmm that definitely helps a bit. I can see how if I split it into 2 cases, 1 where they're equal and one where they're not, you will get it to work for the first case. However the second is not as apparent to me. I seem to get this:

(a - b)(a² + ab + b²) = -(a - b)

(a² + ab + b²) = -1

a² + b² = -(ab + 1)

unfortunately I don't see how to use this for anything. Perhaps I've just been staring at the question too long. It also may be something to do with my not doing a math course in the last 4 years. I had to look up the difference of cubes to see how a³ - b³ even factors.

I'll show what I've done so far, it's similar to what you posted. I would have posted it before but at the time I posted the original message I had not gotten very far with the question as I did know the factoring I mentioned above was possible.

a³ + a = b³ + b

a³ - b³ + a - b = 0

(a - b)(a² + ab + b²) + (a - b) = 0

(a - b) ( (a² + ab + b²) + 1) = 0

so since this basically says the product of 2 numbers is equal to 0 then one of them must be 0. So assuming a - b is the one that's 0 we'll know that:

a - b = 0

a = b

Assuming the (a² + ab + b²) + 1 is 0 leads to a bit of a problem as there's no way that I can see for that to come to 0 since a² + ab + b² cannot come to -1. As for how to prove that formally...not to sure.

Anyhoo I hope the helps you see more where my thinking is.
 
From what you have done, it is very clear that a & b must have the same sign.
Do you see that?
\(\displaystyle \begin{array}{l}
a^3 - b^3 = - (a - b) \\
\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) + \left( {a - b} \right) = 0 \\
\left( {a - b} \right)\left[ {a^2 + ab + b^2 + 1} \right] = 0 \\
\end{array}\).

Because a & b must have the same sign, clearly \(\displaystyle a^2 + ab + b^2 + 1 > 0\).

Thus the only conclusion from \(\displaystyle \left( {a - b} \right)\left[ {a^2 + ab + b^2 + 1} \right] = 0\) is that \(\displaystyle a = b\)
 
pka,

That's clever!

Even after you pointed out - it took me a bit to convince myself that 'a' and 'b' must have same sign.
 
Well after a bit of thinking about it I believe I see what you mean, I think lol. I'll try to concatenate my thoughts into a bit more formal of a proof later and you guys can feel free to tell me what you think. Thanks a lot for the help, if I have any more questions I'll be sure to talk to you guys again.
 
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