Problems

Recall that $(n+1{)}^2=n^2+2n+1$ and after expansion we get $(n+1{)}^2-(2n+1)=n^2$. Subtract $n(2n+1)$ from both sides $(n+1{)}^2-(2n+1)-n(2n+1)=n^2-n(2n+1)$ and rewrite it as $(n+1{)}^2-(n+1)(2n+1)=n^2-n(2n+1)$.
Now we add $\frac{(2n+1{)}^2}{4}$ to both sides: $(n+1{)}^2-(n+1)(2n+1)+\frac{(2n+1{)}^2}{4}=n^2-n(2n+1)+\frac{(2n+1{)}^2}{4}$.
Factor both sides into square: $((n+1)-\frac{2n+1}{2}{)}^2=(n-\frac{2n+1}{2}{)}^2$.
Now take the square root: $(n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}$.
Add $\frac{2n+1}{2}$ to both sides and we get $n+1=n$ which is equivalent to $1=0$.

Look at the following diagram, depicting how to get an extra cell by reshaping triangle.

Can you find a mistake? Certainly the triangles have different area, so we cannot obtain one from the other one by reshaping.

This problem is often called "The infinite chocolate bar". Depicted below is a way to get one more piece of chocolate from the $5\times 6$ chocolate bar. Do you see where is it wrong?

Consider the following "proof" that any triangle is equilateral: Given a triangle $ABC$, we first prove that $AB=AC$. First let’s draw the bisector of the angle $\mathrm{\angle }A$. Now draw the perpendicular bisector of segment $BC$, denote by $D$ the middle of $BC$ and by $O$ the intersection of these lines. See the diagram

Draw the lines $OR$ perpendicular to $AB$ and $OQ$ perpendicular to $AC$. Draw lines $OB$ and $OC$. Then the triangles, $RAO$ and $QAO$ are equal, since we have equal angles $\mathrm{\angle }ORA=\mathrm{\angle }OQA=90°,$ and $\mathrm{\angle }RAO=\mathrm{\angle }QAO,$ and the common side $AO$. On the other hand the triangles $ROB$ and $QOC$ are also equal since the angles $\mathrm{\angle }BRO=\mathrm{\angle }CQO=90°$, the hypotenuses $BO=OC$ the legs $RO=OQ$. Thus, $AR=AQ,$ $RB=QC,$ and $AB=AR+RB=AQ+QC=AC.$ Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that $AB=BC$ in the same way.

Let’s prove the following statement: every graph without isolated vertices is connected.
Proof We use the induction on the number of vertices. Clearly the statement is true for graphs with $2$ vertices. Now, assume we have proven the statement for graphs with up to $n$ vertices.
Take a graph with $n$ vertices by induction hypothesis it must be connected. Let’s add a non-isolated vertex to it. As this vertex is not isolated, it is connected to one of the other $n$ vertices. But then the whole graph of $n+1$ vertices is connected!

Let’s prove that $1=2$. Take a number $a$ and suppose $b=a$. After multiplying both sides we have $a^2=ab$. Subtract $b^2$ from both sides to get $a^2-b^2=ab-b^2$. The left hand side is a difference of two squares so $(a-b)(a+b)=b(a-b)$. We can cancel out $a-b$ and obtain that $a+b=b$. But remember from the start that $a=b$, so substituting $a$ for $b$ we see that $2b=b$, dividing by $b$ we see that $2=1$.

Let’s prove that $1$ is the smallest positive real number: Assume the contrary and let $x$ be the smallest positive real number. If $x>1$ then $1$ is smaller, thus $x$ is not the smallest. If $x<1,$ then $\frac{x}{2}<x$ so $x$ can not be the smallest either. Then $x$ can only be equal to $1$.

In how many ways can you read the word TRAIN from the picture below, starting from T and going either down or right at each step?

There are $100$ people in a room. Each person knows at least $67$ others. Show that there is a group of four people in this room that all know each other. We assume that if person $A$ knows person $B$ then person $B$ also knows person $A$.

The numbers $a$ and $b$ are integers and the number $p\ge 3$ is prime. Suppose that $a+b$ and $a^2+b^2$ are divisible by $p$. Show that $a^2+b^2$ is divisible by $p^2$.