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seen Dec 7 at 15:09

"You have enemies? Good. That means you've stood up for something, sometime in your life" - Winston Churchill


May
6
comment $n^2 + 3n +5$ is not divisible by $121$
You got a great answer by Bill, when a number is divisible by $121$ what is it congruent to mod $11^2$?
May
6
comment The maximum number of nodes in a binary tree of depth $k$ is $2^{k}-1$, $k \geq1$.
@rajansthapit what is the name of tile/author of the book? Gadi has the correct answer.
May
6
answered Derivative of $e^{-x}$
May
6
comment The Importance Of Good Teachers and Guidance In the Academics
Today at least there are tons of resources as an alternative, assuming you are self-motivated. If a teacher is not good, it really brings down the confidence of a student (Personally from what I noticed in my Son's case).
May
6
comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
My +1 and Thanks.
May
6
comment If $b^2$ is the largest square divisor of $n$ and $a^2 \mid n$, then $a \mid b$.
@BillDubuque Recommend the long line to be split as $$ \begin{align*} a^2 | b^2c \Rightarrow a^2 | (bc)^2 &\Rightarrow a^2 | a^2c, abc, b^2c \\ &\Rightarrow a^2 | (a,b)^2c \Rightarrow (a/(a,b))^2 | c \end{align*} $$ My +1 for this answer.
May
6
comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$
@PaulManta this converges to $0.89$, but you have to think how to show that.
May
6
comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$
@PaulManta I changed the \cdots to \times because it was confusing with the numbers themselves with decimal values.
May
6
revised Result of the product $0.9 \times 0.99 \times 0.999 \times …$
edited title
May
6
comment Result of the product $0.9 \times 0.99 \times 0.999 \times …$
How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i}) $
May
6
awarded  Necromancer
May
5
awarded  Nice Answer
May
5
comment Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$
@N3buchadnezzar your expected answer has an error. Checked against Wolfram Alpha's snipurl.com/23dq2hp and instead of plus sign, it shows minus sign. Follow J.M's hint, your might approach the answer.
May
5
comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
@dato, the original question was prove $1-xy$ is a perfect square, so there is a proof required, not solutions to $x,y$.
May
5
comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
I think I do understand both your comments. Bill likes to welcome new users, and Patrick was right to mention that people who seek help should not just post a question without proper introduction. I think MSE should stop unregistered users to just post a question and vanish in thin air.
May
5
comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
@PatrickDaSilva Thanks.
May
5
answered If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
May
5
comment If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.
Did you at least find trivial ones, like $x=y=0$ then $1-xy=1=1^2$ ?
May
5
answered Funny identities
May
5
answered Integration: area enclosed by graph of $x^4 + y^4 = 1$