Solve a system of integers

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Solve a system of integers

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#1 h Mar 30, 2025, 12:08 PM

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Find all quadruples $a,b,c,d$ where $a,b,c,d$ are all integers and $abcd\neq 0$ such that $\begin{cases} a^2+b^2+c^2+d^2\leq 400.\\ a^3+b^3+c^3+d^3\geq 4000. \\ a^4+b^4+c^4+d^4\leq 40001. \end{cases}$

This post has been edited 1 time. Last edited by nhathhuyyp5c, Mar 30, 2025, 3:48 PM
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no_room_for_error

#2 h Mar 30, 2025, 2:03 PM • 1 Y

Y by nhathhuyyp5c

nhathhuyyp5c wrote:

Find all triplets $a,b,c,d$ where $a,b,c,d$ are all integers and $abcd\neq 0$ such that $\begin{cases} a^2+b^2+c^2+d^2\leq 400.\\ a^3+b^3+c^3+d^3\geq 4000. \\ a^4+b^4+c^4+d^4\leq 40001. \end{cases}$

It's called a quadruplet when there are 4 of. (Although I think in math, triple/quadruple is used more - forgive my ignorance, I am not a native speaker). Notice that

$1 \geq a^2(a-10)^2+b^2(b-10)^2+c^2(c-10)^2+d^2(d-10)^2\geq 0$
If $\sum a^2(a-10)^2 =0$ , we get $a=b=c=d=10$ which is a solution.
If $\sum a^2(a-10)^2 =1$ , we get $\sum a^2=400$ and $\sum a^4=40001$ , which is not possible since $a^2\equiv a^4 \pmod 2$ .

This post has been edited 1 time. Last edited by no_room_for_error, Mar 30, 2025, 2:03 PM

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