Solutions to a diophantine equation

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Find the number of ordered quadruples of non-zero integers $(a, b, c, d)$ which satisfy the following equation: $\frac{a^3}{bcd} + \frac{b^3}{cad} + \frac{c^3}{abd} + \frac{d^3}{abc}= 5$

The answer is 0.

2 solutions

Jubayer Nirjhor
Dec 15, 2013

The given expression simplifies to... $a^4+b^4+c^4+d^4=5abcd$

Let $f_{b,c,d} (a)=a^4+b^4+c^4+d^4$ . Note that, $f_{b,c,d}(a)$ is always positive and $(a,b,c,d)\in \mathbb{Z}-\left\{0\right\}$ .

$f_{1,1,1}(1)=4<5<f_{1,1,1}(2)=19$

Hence, $f_{b,c,d}(a)\neq 5$ for any $(a,b,c,d)\in \mathbb{Z}-\left\{0\right\}$ .

The answer is $\fbox{0}$ . :p

Sorry, the proof is wrong... At the last part I solved for $f_{b,c,d}(a)=5$ instead of $5abcd$ . :p

Here's another approach... The given expression simplifies to... $a^4+b^4+c^4+d^4=5abcd$ $5$ has $2$ quartic residues: $0,1$ . The right side is divisible by $5$ , hence so should be the left side. So this is obvious that each of $a,b,c,d$ is divisible by $5$ . Let's write them as $5w,5x,5y,5z$ respectively. Thus we have... $(5w)^4+(5x)^4+(5y)^4+(5z)^4=5\cdot 5w\cdot 5x \cdot 5y\cdot 5z$ $\Longrightarrow ~~~ 5^4(w^4+x^4+y^4+z^4)=5^5 wxyz ~~~~\Longrightarrow ~~~w^4+x^4+y^4+z^4=5wxyz$

Let's assume that $(a_1, b_1, c_1, d_1)$ are the values of $(a,b,c,d)$ respectively such that the minimum of $a+b+c+d$ is attained. Now since $a,b,c,d$ are non-zero, $(w,x,y,z)<(a,b,c,d)$ , implying $w+x+y+z<a+b+c+d$ , which is a contradiction since $(w,x,y,z)$ is also a solution to the original equation.

Hence, no solutions exist to the given equation. Answer: $\fbox{0}$ :)

Jubayer Nirjhor - 7 years, 5 months ago

Nice! This is moreorless my intended solution. The trick you used is also known as Infinite Descent. However, you are missing a small detail. You are taking it for granted that $(a, b, c, d)$ are positive. That can easily be fixed though, by noting that $5abcd>0$ , so the negative values come in pairs, which essentially means $(a, b, c, d)$ satisfies the equation if and only if so does $(|a|, |b|, |c|, |d|)$ .

Sreejato Bhattacharya - 7 years, 5 months ago

Yes, I missed that. Thanks! :D

Jubayer Nirjhor - 7 years, 5 months ago

A Brilliant Member
Dec 16, 2013

This reduces to $a^4+b^4+c^4+d^4 = 5abcd$ . For any integer $n$ , $n^4 \equiv 0,1 \pmod{5}$ . There are four numbers in LHS, & any combination of four 0s or 1s doesn't make 5. So each $a \equiv b \equiv c \equiv d \equiv 0 \pmod{5} \Rightarrow a=5a_1 \cdots$ same for others.

We again arrive at $a_1^4 + b_1^4+c_1^4+d_1^4=5a_1b_1c_1d_1 \Rightarrow a_1=5a_2 \cdots$ etc. Thus, by infinite descent it follows there are no non-zero solutions. [Concept of infinite descent: $a=5a_1 =5^2a_2=5^3 a_3= \cdots = 5^na_n$ , as $n$ tends to infinity this becomes impossible for some non-zero integer.]

Solutions to a diophantine equation

The answer is 0.

2 solutions

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