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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??

Please help me. Thanks in advance,

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

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

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

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