# Solving Diophantine equation with parameters

Today i want to learn/discuss about of a technique for proving that the given Diophantine equation has infinitely main roots.

I just want to solve these types of problems : 1) Show that $$x^2=y^3+z^5$$ has infinitely many solutions for positive integers $$x,y,z$$. 2) Show that $$x^n+y^n=z^{n-1}$$ has infinitely many solutions for postive integers $$x,y,z$$.

To solve these types of problems a parameter is usually used which varies over integers giving infinitely many solutions.. For example, $$x=k^{10}(1+k)^8 ,y=k^7(1+k)^5,z=k^4(k+1)^3$$ are the solutions for problem number (1). Then as we vary $$k$$ over postive integers we will be getting infinitely many values.

My problem is that I am not getting motivation to how to select that values of $$x,y,z$$ in terms of $$k$$ or any parameter. After seeing the solution I feel "Ohk! It can be done using this" but i can't predict the solution. Hence I am asking is there any procedure are to be followed to solve these types of problems , any motivation inside the question?? Or it can be solved only by putting random value??

• One technique is to find a couple solutions or have a computer find them and then factor each solution to look for commonalities. – abiessu Oct 2 at 12:31

"OP" inquired about, $$x^n+y^n=z^{n-1}$$

For n=3, above equation becomes:

$$x^3+y^3=z^2 ----(1)$$

Equation $$(1)$$ has parametric solution given below:

$$x=2(m^2+1)(m^4+3)$$

$$y=2(m^2-1)(m^4+3)$$

$$z=4m(m^4+3)^2$$

For, $$m=3$$, we get:

$$1680^3+1344^3=(84672)^2$$

• I want to find the general method of thinking but not the solution. – Sufaid Saleel Oct 3 at 3:14

One of my friend on Aops @Delta0001 has discovered this method for solving these types of methods of one type. So I want to share that trick to the math community.

Prove that there are infinite tuples of natural numbers $$(a_1 , a_2 , \dots , a_n)$$ which satisfy

$$a_1 ^{p_1} + a_2 ^{p_2} + \dots + a_n ^ {p_n} = d^p$$

where $$p , p_1, p_2 , \dots , p_n , d$$ are natural numbers and $$gcd ( ~ lcm( p_1 , p_2 , \dots , p_n ) ~ , ~ p) = 1$$

Solution

Let $$M$$ denote the LCM of $$p_1 , p_2 , \dots , p_n$$.

Next, we find $$k$$ such that $$M \cdot k + 1$$ is a multiple of $$p$$.

Now, take any natural numbers $$b_1 , b_2 , \dots , b_n$$ and set,

\begin{align*} a_1 &= b_1 (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_1}} \\ a_2 &= b_2 (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_2}} \\ & \vdots \\ a_n &= b_n (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_n}} \end{align*}

Also, let $$d = (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k + 1}{p}}$$

It is clearly seen that this tuple $$a_1 , a_2 , \dots , a_n , d$$ satisfy the given equation.

And as there are infinitely many choices for $$b_1 , b_2 , \dots m b_n$$, we get infinitely many solutions. $$\blacksquare$$