HKDSE Mathematics Question Review (20241114))

Question:

Given x and p are integers, show that:

p^2 + 1 = x (x+1) (x+2) (x+3)

without using mathematical induction.

Note that both x and p are positive.

p^2 - 1 = x (x+1) (x+2) (x+3)

Solution:

P+1 = (x+1)(x+3) = ((x+2)+1) ((x+2)-1)

P-1 = x (x+2) = ((x+1)+1) ((x+1)-1)

Since x is an integer, p must also be integer.

Further factorization will yield surds.

Note: P+1 = (q + i) (q -i)

Where i^2= -1 and q^2=p

So basically you can observe 2 complex factors.

Further observation:

Not all polynomials can be factorized in radical forms, especially when the degree is 5 or above. For example:

x^5 + x^4 + x^3 + x^2 + i

You can factorize this polynomial because the product of roots -i is not commutative.

You can't factorize this polynomial because the product of roots -i is not commutative.

More:

Further reading:

Sorry, the proof is not accurate enough.

Without loss of generality:

x(x+1)(x+2)(x+3) = p^2 -1

Since p^2 is a perfect square,

p must be an integer

Since the left hand side is commutative,

we introduce an arbitrary constant k >0 ,

such that:

p+1 = k(x+1)(x+3)

p-1 = x(x+2)/k

Now we put x=1, then:

1•2•3•4=24=4•6=5^2-1

P-1 = 4 = 3/k

k= 3/4

P+1 = 2 • 4 • 3/4 = 6

QED

P must be a positive integer

Recall P+1:

Wec can say that every real number is the product of 2 conjugate complex numbers.

For example, take P + 1 =10, then we can factorize it into 2 complex numbers:

P as an integer =9 = (q+i)(q-i) = 10 - 1

Then q^2=10

q must be a surd.

We can say that every real number is the product of 2 conjugate complex numbers.

Now, consider (q+r)(q-r) = P + 1

then for r^2= -s <0,

r, s as arbitrary positive real numbers,

we can exhaust:

P + 1 = 41 = 25 + 16

then q = 5 and r=4i for s=16

Or we can define

(r+q) (r-q) = P +1 for q^2 = -t <0

such that q = 5i and r=4 for t=25

That's why real numbers are always the subset of complex numbers.