Number theory example problems

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Q.
$a^4 + a^2 + 1$ is a prime number
Find the number of possible values of $a$

(i) if $a \in \mathbb{N}$
(ii) if $a \in \mathbb{I}$

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Solution:

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$$a^4 + a^2 + 1 = (a^2)^2 + 1^2 + 2a^2 = (a^2 + 1)^2 - a^2$$ $$a^4 + a^2 + 1 = (a^2 + 1 + a)(a^2 + 1 - a)$$

So,

$$a^4 + a^2 + 1 = (a^2 + 1 + a)(a^2 + 1 - a)$$

→ Since these are 2 factors of a prime number, one of them is 1 and the other is the prime number itself.

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Case I: First factor is 1.

$$a^2 + a = 0 \Rightarrow a(a + 1) = 0 \Rightarrow a = 0, a = -1$$

Now check both in $a^2 + 1 - a$:

• $a = 0$: $0^4 + 0^2 + 1 = 1$ → Not prime

• $a = -1$:

  $$1 + 1 + 1 = 3 \Rightarrow \text{prime}$$

✅ So only valid value is $a = -1$

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Case - II

$$a^2 + 1 - a = x \Rightarrow a^2 - a = 0 \Rightarrow a(a - 1) = 0 \Rightarrow a = 0, a = 1$$

Now check in $a^2 + 1 + a$ if it's prime:

• $a = 0$: Not prime

• $a = 1$:

  $$1 + 1 + 1 = 3 \Rightarrow \text{prime} \quad \checkmark$$

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Q2)
For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?

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$$n^2 - 3n + 2 = (n - 1)(n - 2)$$

Now, test values:

Case ①

$$n - 1 = 1 \Rightarrow n = 2 \\ n - 2 = 0 \Rightarrow \text{Not a prime} \quad ✗$$

Case ②

$$n - 2 = 1 \Rightarrow n = 3 \\ n - 1 = 2 \Rightarrow \text{Prime} \quad \checkmark$$

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So only one positive integer $n = 3$ makes the expression prime.
But we forgot to check if one factor is -1 and other factor is negative of the prime.

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(iii)

$$n - 1 = -1 \Rightarrow n = 0 \quad ✗ \\ (n - 2) \text{ is negative} \\ |n - 2| \rightarrow \text{prime} \quad \text{✗ not valid}$$

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(iv)

$$n - 2 = -1 \Rightarrow n = 1 \\ n - 1 \rightarrow \text{negative} \Rightarrow \text{✗}$$

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Final Answer:

$$\boxed{\text{Ans: } 1}$$

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Q.
$a, b, c$ are prime numbers

$$a + b + c = 66$$ $$ab + bc + ca = 1071$$

Find:

$$abc = \ ?$$

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Out of a,b,c one is even and rest 2 are odd. No other combination is possible.
Only even prime is 2.
WLOG, a = 2.
This gives 2 equations in 2 variables (b,c).
You will get a quadratic solving which you will get b,c = 23,41

Q.
The number of positive integral values of $n$
for which $n^3 - 8n^2 + 20n - 13$ is a prime number.

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Solution:
Using rational root theorem we can identify one of the roots as 1.
Dividing by (n-1) we get the other factor as (n^2 - 7n + 13).

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Case – I

$$n - 1 = 1 \Rightarrow n = 2$$

Now check:

$$n^2 - 7n + 13 \Rightarrow 4 - 14 + 13 = 3 \quad \text{✓ prime}$$

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Case – II

$$n^2 - 7n + 13 = 1$$ $$n^2 - 7n + 12 = 0 \\ n^2 - 3n - 4n + 12 = 0 \\ (n - 3)(n - 4) = 0 \\ n = 3, 4$$

Now check whether $n - 1$ is prime.
n-1 = 2,3 So valid.

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Q.
The number of natural numbers $n$ for which

$$\frac{15n^2 + 8n + 6}{n}$$

is a natural number.

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Breakdown:

$$\frac{15n^2}{n} + \frac{8n}{n} + \frac{6}{n} = 15n + 8 + \frac{6}{n}$$

For the expression to be a natural number:

$$\frac{6}{n} \text{ must be a natural number} \Rightarrow n \text{ must be a factor of } 6 \Rightarrow n = 1, 2, 3, 6$$

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Other values of n can be -1, -2, -3, -6 but then the original number wouldn't remain a natural number.

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Q.
Find the number of integer $n$ for which

$$\frac{n}{20 - n}$$

is a square of an integer.

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Solution 1:
If n  = 0 then it's square of 0.
If n != 0 then 0 < n < 19 and denominator should be less then equal to n.
=> n >= 20 - n => n >= 10
Try n from 10 to 19.
n = 10, 16,18 work.
So answer = 4.

Solution 2:
Write it as:
(n - 20 + 20)/(20-n)
= -1 + 20/(20 - n)
For it to be integer (20-n) should be a factor of 20
=> 20 - n = 1,2,4,5,10,20 Or their negative values.
From that you will get 4 valid values again.

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Q.
$m, n$ are natural numbers.
Find the number of pairs $(m, n)$ for which

$$m^2 + n^2 + 2mn - 2013m - 2013n - 2014 = 0$$

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Solution:
Let m+n = x then it's a quadratic in x with roots = -1, 2014
Since m,n are natural -1 is invalid.
Number of ordered pairs = 2013.

Q.
For how many natural numbers $n$ between 1 and 2014
(both inclusive) is

$$\frac{8n}{9999 - n}$$

an integer?

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Solution1:
It can be written as:
8/(9999/n - 1)
=> 9999/n - 1 is a factor of 8.
Try putting the values 1,2,4,8 and their negatives.
Only integer values of n you will get are 3333,1111 but the first one is > 2014. So only 1 answer.

Sure! Here's the converted text from the image:

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Q. Let $S$ be the set of all integers $n$ such that

$$\frac{8n^3 - 96n^2 + 360n - 400}{2n - 7}$$

is an integer.

Find the value of

$$\sum_{n \in S} |n|$$

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Solution outline

1. Polynomial-long division

$$\frac{8n^{3}-96n^{2}+36n-400}{2n-7}=4n^{2}-34n-101-\frac{1107}{2n-7}.$$

So the fraction is integral ⇔ $\displaystyle\frac{1107}{2n-7}$ is integral.

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2. Divisibility condition

This means $2n-7\mid 1107$.

$$1107=3^{3}\cdot 41 \quad\Longrightarrow\quad \text{all divisors }d=\pm1,\pm3,\pm9,\pm27,\pm41,\pm123,\pm369,\pm1107.$$

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3. Corresponding integer $n$

For each divisor $d$,

$$2n-7=d\;\Longrightarrow\; n=\frac{d+7}{2}.$$

Because every $d$ is odd, $d+7$ is even, so each gives an integer $n$.

$d$	$n=\dfrac{d+7}{2}$
$\pm1$	4, 3
$\pm3$	5, 2
$\pm9$	8, –1
$\pm27$	17, –10
$\pm41$	24, –17
$\pm123$	65, –58
$\pm369$	188, –181
$\pm1107$	557, –550

Thus

$$S=\{4,5,8,17,24,65,188,557,3,2,-1,-10,-17,-58,-181,-550\}.$$

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4. Required sum

$$\sum_{n\in S}\lvert n\rvert =4+5+8+17+24+65+188+557+3+2+1+10+17+58+181+550 =1690.$$

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$$\boxed{{\displaystyle 1690}}$$