1,912 reputation
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location Germany
age 28
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2d
awarded  Yearling
Jul
6
comment Solving congruences
@JoeJohnson126 because 7 is not equivalent to 1 mod 5 (congruence II)
Jul
6
comment Solving congruences
Thanks, with necessary/sufficient it seems quite logical. Thanks a lot.
Jul
6
accepted Solving congruences
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
28
asked Solving congruences
May
21
comment Problem with root equation
What have you tried? Use $\sqrt{4-x^2}=\sqrt{(2-x)(2+x)}$ To control your work: the solution ist $x=\frac 6 5$
Apr
24
accepted Uniqueness of Point
Apr
3
comment Uniqueness of Point
and how do I argue for $X \notin [PQ]$ and the 3rd case $a=b$?
Apr
2
answered Uniqueness of Point
Apr
2
comment Uniqueness of Point
I have to plug in $d(X,P) + d(X,Q) = |P-Q|$ for e.x. $d(X,P)$ or how should I start?
Apr
2
asked Uniqueness of Point
Mar
27
revised Ring theory question: $I=\langle x,2 \rangle$ prime/maximal ideal in $\mathbb Z[x]$?
improved formatting
Mar
27
suggested suggested edit on Ring theory question: $I=\langle x,2 \rangle$ prime/maximal ideal in $\mathbb Z[x]$?
Mar
27
comment Ring theory question: $I=\langle x,2 \rangle$ prime/maximal ideal in $\mathbb Z[x]$?
Anything you tried?
Mar
6
awarded  Nice Answer
Mar
2
accepted Proof: $\lambda = \frac 1 2 (d(X,A)^2-d(X,B)^2+1)$ for $d(A,B)=1$
Mar
2
comment Proof: $\lambda = \frac 1 2 (d(X,A)^2-d(X,B)^2+1)$ for $d(A,B)=1$
Oh, that was helpful. I tried to use all the time $x=a+\lambda(b-a)$ and got stuck every time. Thanks anyway.
Mar
2
comment Proof: $\lambda = \frac 1 2 (d(X,A)^2-d(X,B)^2+1)$ for $d(A,B)=1$
thanks, I edited it.