Consider the equation x^2+y^5=z^3 where x, y, and z, are positive integers.?

Answer 1

There are infinite solutions to this equation in positive integers.

x = 2^a ; y=2^b ; z=2^c

a, b, c positive integers and
where 2a = 5b & 2a+1 = 3c
will yield a solution.

[ x^2 + y^5
= (2^a)^2 + (2^b)^5
= 2^(2a) + 2^(5b)
= 2^(2a) + 2^(2a) = 2*2^(2a) = 2^(2a+1) = 2^(3c) = z^3 ]

so any
a = any multiple of 5 that is 1 mod 3 will work.
i.e., a=10, b=4, c=7 yields x=2^10 = 1024; y=2^4 = 16; z=2^7 = 128 works, since 1024^2 + 16^5 = 1048576 + 1048576 = 2097152 = 128^3



{ restated,
For n = 1, 2, 3, 4, ...

x = 2^[5(3n - 1)]
y = 2^[2(3n - 1)]
z = 2^(10n - 3)
}

Answer 2

of course y and z can not the two be even using fact this makes LHS even and RHS surprising. Can y, z be equivalent? if so then x*y^2 = 2y^2 + a million ----> (x - 2)*(y^2) = a million and clearly x = 3, y = a million is the only answer with integers. any extra i will assume that y is the better of y, z as reversed recommendations are particularly a similar. Can z = a million be a answer with out y = a million? xy = y^2 + 2 ----> y^2 - xy + 2 = 0 ----> y = [x + sqrt(x^2 - 8)]/2 x = 3 finally ends up in y = 2 so x = 3, y = 2, z = a million is a answer with all 3 diverse. Can z = 2? 2xy = y^2 + 5 ----> y^2 - 2xy + 5 = 0 ----> y = [2x + sqrt(4x^2 - 20)]/2 = x + sqrt(x^2 - 5) This has answer x = 3, y = 5, z = 2. For sqrt(x^2 - 5) to be an integer we want x^2 - 5 = ok^2 yet 2^2 and 3^2 are the only squares differing by making use of 5 so this finally ends up in no different recommendations. I even have discovered, yet won't placed the evidence right here, that z (the smaller of y, z bear in recommendations) can't be 3 or 4. This makes me suspect that there at the instant are not the different recommendations. EDIT. That final line is misguided! i've got only discovered that x = 3, y = 13, z = 5 is a answer. in spite of the fact that, this is calling as though x must be 3.

Answer 3

ill start:
x^2=z^3 - y^5,
find value of x. x = sq. root of (z^3 - y^5,)

subst. to eq. then find y.

youll get
y^5= z^3 - z^6

dude, thats a long way.... hehe. im better at pen and paper than on typing equations.

Answer 4

omg you must be kidding! lol I didnt excell in math sorry!

Answer 5

plz elaborate

Answer 6

what's the question?

Answer 7

Could you please elaborate?