# How do I solve x^3 + y^3 = 1, x^4 + y^4 = 1

1. Feb 27, 2005

### mohlam12

hey everyone, here is my problem, u have to solve this:
x^3 + y^3 = 1
x^4 + y^4 = 1

i tried to solve it but i couldnt, please just give me some hints if you can, thx

i had this one too,
2x+xy+2y = 59
3x-2xy+3y = -34

the teacher told me to do:
2(x+y)+xy=59
3(x+y)-2xy=-34

then,
x+y=12
xy=34

then i solved the equation : x^2-12x+35=0
i got x=5 and x=7
so, the solutions are
x=5 y=7 and x=7 and y=5

i didn t get it, he explained to me but i still don t get why, can anyone explain to me how it was solved???

Last edited: Feb 27, 2005
2. Feb 27, 2005

### dextercioby

For the first,here's what my maple had to say:
Solution is : $\left\{ y=0,x=1\right\} ,\allowbreak \left\{ y=1,x=0\right\} ,\allowbreak \left\{ y=-2\rho ^4-4\rho ^3-6\rho ^2-5\rho -3,x=\rho \right\}$ where $\rho$ is a root of $2Z^6+6Z^5+12Z^4+16Z^3+15Z^2+9Z+4$

,which,translated,means:
Solution is $$\{y=0,x=1\},\{y=1,x=0\},\{y=-2\rho^{4}-4\rho^{3}-6\rho^{2}-5\rho-3,x=\rho\} \where \ [\tex] [\tex] \rho \ is \ a \ root \ of \ 2z^{6}+6z^{5}+12z^{4}+16z^{3}+9z+4$$

Daniel.

3. Feb 27, 2005

### dextercioby

Yes,those substitutions turn the initial system into one which is simpler.There's not too much to explain,though,these things are working only with particular examples,you can generalize in any way.

It's more of an inspiration,really.You either have it (case in which u see the trick you have to pull),or not.

Daniel.

4. Feb 27, 2005

### mohlam12

and...do u know how to get {0,1} {1,0} ... without ur maple ???

5. Feb 27, 2005

### dextercioby

You can see it directly that,y chosing x=0,then "y" is a solution of the system:
$$y^{3}=1;y^{4}=1$$,which is of course y=1.
The same if you chose y=0,you'll find x=1,that's because the system is homogenous in "x" and "y".

Daniel.

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