Question

Algebra SUCKS !!

Answer 1

\[\large Let\]
\[\large x^3-y^2=z^5\]
\[\large x+z=y\]

where x,y and z are all integers less than 19.

Find x+y+z.

Answer 2

@ganeshie8 and @PeterPan

Answer 3

I don't know this stuff. :|

Answer 4

You can always make stuff up, say, 2? :)

Answer 5

@ganeshie8 any idea?

Answer 6

The only thing I think might be worth mentioning is that:

x+y+z=2y

That might be useful, but honestly I don't know.

Answer 7

Just let z = 0 and x = 1
and everything falls into place XD

Answer 8

Seems good enough to me.

Answer 9

@eliassaab and @CGGURUMANJUNATH

Answer 10

x+y+z<= 57

Answer 11

There is only one situation to achieve that
x=7
y=10
z=3
so x+y+z=20

Answer 12

if you assume all less than 19

Answer 13

z^5<6498
z<=5

Answer 14

actually
x+y+z = 2y < 2*19 = 38

so x+y+z < 38 @ikram002p

Answer 15

Here is the program in Mathematica that gives you the answer
Clear[x, y, z]
Q = {};
For[ x = 1, x < 19, x++,
For[ z = 1, z < 19, z++,
y = x + z;
If [x^3 - y^2 == z^5, Q = Append[Q, {x, y, z}]]]]
Q

Answer 16

wow ! u have something working !xD

Answer 17

typo :o

Answer 18

If you add 0, then here are the three solutions
{{0, 0, 0}, {1, 1, 0}, {7, 10, 3}}

Answer 19

Clear[x, y, z]
Q = {};
For[ x = 0, x < 19, x++,
For[ z = 0, z < 19, z++,
y = x + z;
If [x^3 - y^2 == z^5, Q = Append[Q, {x, y, z}]]]]
Q

Answer 20

@eliassaab can you please explain me how do you get your answer?

Answer 21

The computer found them. I only wrote the script.

Answer 22

what is the site

Answer 23

Even, when doing by hand, you have to do trial and error after you narrow down the
possibilitiess

Answer 24

The site is my computer and Mathematica working on it

Answer 25

(0, -1, -1) is one solution

mayank, we need to find x+y+z in in all integers
or only positive ?

Answer 26

if it is all integers,
then clearly x+y+z will not have an unique answer

Answer 27

Im the smartest man alive jk lol

Answer 28

I found the same answer, if all are less than 1000

Answer 29

@ganeshie8 explain your solution

Answer 30

ganesh telling u is the solution in all integer or in +ve

Answer 31

I am asking if the domain for x, y, z is < 19
or only the positive integers < 19 ?

Answer 32

x+y-z<19
x+y+z<38
z<=5
x<=7

Answer 33

"Teacher I made this program since there are only 19^3 possibilities..."

Answer 34

i cant see any method exept trial :o

Answer 35

me neither,
wid ur bounds we can go trial and error i feel

Answer 36

Fun fact, if you change thist equation to lower each term by 1 degree to:

x^2-y=z^4

Then you get:
x=2, y=3, z=1 so then you have x+y+z=6.

whooo...

Answer 37

can you explain? @Kainui

Answer 38

but it leads to nothing lol

Answer 39

@ikram002p do you know how to solve it?

Answer 40

trials method see eliassaab methos is like this
set
x=0,y=0,z=0
s,t
all x,y,z<19
then check if
y=x+z
then
x^3-y^2=z^5
and keep going

Answer 41

As I said before the following program spit them all
Clear[x, y, z] Q = {}; For[ x = 0, x < 19, x++,
For[ z = 0, z < 19, z++,
y = x + z;
If [x^3 - y^2 == z^5, Q = Append[Q, {x, y, z}]]]]
Q
and here are all the solutions between and including zero to 19
{{0, 0, 0}, {1, 1, 0}, {7, 10, 3}}
When you want to do trial do it by the machine. It is faster.