(0,0,0,0)=(a,b,c,d)? Then 0+0+0-0n=0 so n=1 works I'm kinda confused by the wording

PEKKA wrote: (0,0,0,0)=(a,b,c,d)? Then 0+0+0-0n=0 so n=1 works I'm kinda confused by the wording There are other solutions to $a^2 + b^2 + c^2 - d^2 = 0$, so $n=1$ doesn't work.

hood09 wrote: find the smallest integer $n\geq1$ such that the equation : $$a^2+b^2+c^2-nd^2=0 $$has $(0,0,0,0)$ as unique solution . for $n=1$ take $a=b=2$ and $c=1$ ,$d=3$ for $n=2$ take $a=b=1$,$c=4$,$d=3$ for $n=3$ take $a=b=c=d=1$. for $n=4$ fermat infinity desert with $4$ gives $a=b=c=d=0$ the only solution.(wrong because after the.forst time i lost the $4$ on front of $d$)

For $n=4 \to$ $a=b=4$ and $c=2,d=3$ is a solution (basically analogous to case $n=1$). So $n=4$ is not the answer.

It's actually $n=7$. For $n\leq6$ just find $(a,b,c,d)$ such that $(a,b,c,d) \neq (0,0,0,0)$ For $n = 7$ the equation becomes $a^2+b^2+c^2=7d^2$ Just consider $mod \ 8$ and then infinite decent on $a, b, c, d$ we get that $(a,b,c,d) = (0,0,0,0)$ is the only solution

Tang_Tang wrote: It's actually $n=7$. For $n\leq6$ just find $(a,b,c,d)$ such that $(a,b,c,d) \neq (0,0,0,0)$ For $n = 7$ the equation becomes $a^2+b^2+c^2=7d^2$ Just consider $mod \ 8$ and then infinite decent on $a, b, c, d$ we get that $(a,b,c,d) = (0,0,0,0)$ is the only solution For $n=5 \Rightarrow a=5,b=4,c=2, d=3$ For $n=6 \Rightarrow a=2,b=1,c=1,d=1$

P2nisic wrote: hood09 wrote: find the smallest integer $n\geq1$ such that the equation : $$a^2+b^2+c^2-nd^2=0 $$has $(0,0,0,0)$ as unique solution . for $n=1$ take $a=b=2$ and $c=1$ ,$d=3$ for $n=2$ take $a=b=1$,$c=4$,$d=3$ for $n=3$ take $a=b=c=d=1$. for $n=4$ fermat infinity desert with $4$ gives $a=b=c=d=0$ the only solution.(wrong because after the.forst time i lost the $4$ on front of $d$) if n=4 then (6;8;0;5) is solution