A polynomial $f ( x )$ has integer coefficients such that $f ( 0 )$ and $f ( 1 )$ are both odd numbers. Prove that $f ( x ) = 0$ has no integer solutions.

If 8 points in a plane are chosen to lie on or inside a circle of diameter 2 cm then show that the distance between some two points will be less than 1 cm.

Using the fact that $\sqrt { n }$ is an irrational number whenever $n$ is not a perfect square, show that $\sqrt { 3 } + \sqrt { 7 } + \sqrt { 21 }$ is irrational.

(a) Show that the area of a right-angled triangle with all side lengths integers is an integer divisible by 6.
(b) If all the sides and area of a triangle were rational numbers then show that the triangle is got by 'pasting' two right-angled triangles having the same property.

Show that $\frac { \ln ( 12 ) } { \ln ( 18 ) }$ is irrational.

a) We want to choose subsets $A _ { 1 } , A _ { 2 } , \ldots , A _ { k }$ of $\{ 1,2 , \ldots , n \}$ such that any two of the chosen subsets have nonempty intersection. Show that the size $k$ of any such collection of subsets is at most $2 ^ { n - 1 }$. b) For $n > 2$ show that we can always find a collection of $2 ^ { n - 1 }$ subsets $A _ { 1 } , A _ { 2 } , \ldots$ of $\{ 1,2 , \ldots , n \}$ such that any two of the $A _ { i }$ intersect, but the intersection of all $A _ { i }$ is empty.

We consider a real $\alpha$ such that for every prime number $p$, $p^\alpha$ is a natural number. We apply relation $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f(x+j) = f^{(n)}(x + y_n) \quad \text{(IV.1)}$$ to the function $f_\alpha(x) = x^\alpha$ and to the integer $n = \lfloor \alpha \rfloor + 1$ (where $\lfloor \cdot \rfloor$ denotes the floor function). We now choose $x \in \mathbb{N}^*$.
Show that the expression $$\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} f_\alpha(x+j)$$ is a relative integer.

Show that $n^4 + 4^n$ is composite for all integers $n > 1$.

A natural number $n$ is said to be a perfect square when there exists a natural number $x$ satisfying $n = x ^ { 2 }$. Similarly, $n$ is said to be a perfect cube when there exists a natural number $x$ satisfying $n = x ^ { 3 }$.
In the following two cases, find the natural number $n$ that satisfies the conditions.
(i) $n$ is a perfect square. Furthermore, the number obtained by adding 13 to $n$ is also a perfect square.
(ii) $n$ is a perfect cube. Furthermore, the number obtained by adding 61 to $n$ is also a perfect cube.
First, consider (i). From the definition of a perfect square number, $n$ can be expressed as $n = x ^ { 2 }$, where $x$ is a natural number. In addition, there exists a natural number $y$ such that
$$x ^ { 2 } + 13 = y ^ { 2 } .$$
Since $x < y$, $y - x = \square \mathbf { J }$ and $y + x = \mathbf { K L }$. It follows that
$$x = \mathbf { M } , \quad y = \mathbf { N } ,$$
and finally that $n = \mathbf { O P }$.
Next, consider (ii). Similar to (i), in (ii), there exists a natural number $x$ such that $n = x ^ { 3 }$, and there also exists a natural number $y$ such that
$$x ^ { 3 } + 61 = y ^ { 3 } .$$
When we solve this equation, we obtain
$$x = \mathbf { Q } , \quad y = \mathbf { R } ,$$
and hence the perfect cube $n = \mathbf{ST}$.

Consider integers $a$ and $b$ satisfying the equation $$14a + 9b = 147. \tag{1}$$
(1) We are to find the positive integers $a$ and $b$ satisfying equation (1).
Since $$14a = \mathbf{A}(\mathbf{BC} - \mathbf{D}b) \text{ and } 9b = \mathbf{E}(\mathbf{FG} - \mathbf{H}a),$$ $a$ is a multiple of $\mathbf{A}$, and $b$ is a multiple of $\mathbf{E}$.
So, when we set $a = \mathbf{A}m$ and $b = \mathbf{E}n$, where $m$ and $n$ are integers, from (1) we have $$\mathbf{I}m + \mathbf{J}n = \mathbf{K}.$$
Since the positive integers $m$ and $n$ satisfying this are $$m = \mathbf{L} \text{ and } n = \mathbf{M},$$ we obtain $$a = \mathbf{N} \text{ and } b = \mathbf{O}.$$
(2) We are to find the solutions $a$ and $b$ of equation (1) satisfying $0 < a + b < 5$.
Since $14 \times \mathbf{N} + 9 \times \mathbf{O} = 147$, from this equality and (1) we have $$14(a - \mathbf{N}) = 9(\mathbf{O} - b).$$
Since 14 and 9 are relatively prime, $a$ and $b$ can be expressed in terms of an integer $k$ as $$a = \mathbf{P} + \mathbf{Q}k, \quad b = -\mathbf{RS}k + \mathbf{T}.$$
Since $0 < a + b < 5$, we have $k = \mathbf{U}$, and we obtain $$a = \mathbf{VW}, \quad b = -\mathbf{XY}.$$