please help!! thank you!

(n+1)² + 1 = n² + 2n + 1 + 1 = n² + 2n + 2

Difference between this and n² + 1 = 2n + 1

Since both are divisible by d, if d = n² + 1 then n² + 2n + 2 is a multiple of n² + 1

(n² + 2n + 2)/(n² + 1) = 1 + (2n + 1)/(n² + 1)

(2n + 1)/(n² + 1) must be a positive integer for n² + 2n + 2 to be a multiple of n² + 1.

Realise that n² > 2n for n > 2, (2n + 1)/(n² + 1) will be a proper fraction. So n has to be 2 or smaller than 2 for (2n + 1)/(n² + 1) to possibly be a whole number

If n = 1,
2n + 1 = 3
n² + 1 = 2
3 ÷ 2 = 3/2, which is not a whole number.

If n = 2,
2n + 1 = 5
n² + 1 = 5
5 ÷ 5 = 1

So only n = 2 will satisfy the condition when d = n² + 1. And thus d = 5.

thank you! does anyone know how to do question 4 and 5?

Smallest possible d is 1 since the positive integers are all divisible by 1. As n is a positive integer, n² + 1 and (n+1)² are also positive integers.

n² can be odd or even. Correspondingly, n² + 1 will be even or odd. This is the same for (n + 1)² and (n + 1)² + 1

The first few numbers of n² + 1or (n+1)² + 1 are :
2 (5 10) 17 26 37 (50 65) 82 101 122 (145 170) 197 226 257 (290 325) 362 401 442 (485 530)


There is an alternating sequence of prime and non-prime numbers (the non prime numbers are even and thus divisible by 2 and are not prime as a result) except that after every 3 numbers, there is a pair of numbers which are divisible by 5.

When you pick a pair that contains a prime, the only possible value of d is 1 since primes are only divisble by themselves and 1. It is the same when you pick a pair where 1 number is a bracket number

Only when you pick both numbers in the same bracket will d be equal to 5.

Alternative explanation for 3 : (better than the previous one I posted )


greatest common divisor is also known as the highest common factor (HCF).

Dividend ÷ divisor = quotient (no remainder in this case)

If n² + 1 is divisible by d, then (n+1)² + 1 is also divisible by d.

When expanded, (n+1)² + 1 = n² + 2n + 2

So the difference between the two
= 2n + 1

This 2n + 1 must be divisible by d as well, in order for (n+1)² + 1 to be divisible by d.


Now, if 2n + 1 is divisible by d, then
(2n + 1)(2n - 1) is divisible by d since it's a multiple of (2n + 1).

(2n + 1)(2n - 1) = 4n² - 1

if n² + 1 is divisible by d, then 4(n² + 1) is divisible by d since it's a multiple of (n² + 1)

4(n² + 1) = 4n² + 4

Since 4n² + 4 and 4n² - 1 are both divisible by d, then their difference is also divisible by d.

Difference = 4n² + 4 - (4n² - 1) = 5

The only factors of 5 are 5 and 1.

If 5 is divisible by d,
then d can only be 5 or 1


The idea here is to get rid of all the terms in n

thank you!