What is the power of x and y ? Picture not very clear.

2 for both

If x and y are odd integers, then :

x = 2k + 1 , k ∈ Z (set of integers)
y = 2n + 1, n ∈ Z

x² = (2k + 1)² = 4k² + 4k + 1
y² = (2n + 1)² = 4n² + 4n + 1


When k = n,

x² + y²

= 2(4k² + 4k + 1)

= 2(2k + 1)²

If x² + y² is a perfect square, then it could be written in the form w² where w ∈ Z


Then 2(2k + 1)² = (√2(2k + 1))²

2k + 1 is an integer but √2 is irrational so √2(2k + 1) is irrational.

Which means that it cannot be written in the form w² where w ∈ Z, so it cannot be a perfect square.

When x ≠ y,

x² + y²

= (2k + 1)² + (2n + 1)²

= 4k² + 4k + 1 + 4n² + 4n + 1

= 4(k² + k + n² + n) + 2


If we divide (x² + y²) by 4, it leaves a remainder of 2.



However for perfect square numbers,


Odd square numbers are in the form (2m + 1)²
= 4m² + 4m + 1
= 4(m² + m) + 1, where m ∈ Z

Even square numbers are in the form (2m)²
= 4m²


The former leaves a remainder of 1 when divided by 4 but the latter leaves no remainder.


So x² + y² doesn't belong to either case and so is not a perfect square

To prove :

If n is a composite positive integer, then n ∣ (n - 1)!

Cases :

① n = ab where both a and b are composite themselves

a,b ∈ Z+
a,b < n
a < b

Since (n - 1)! = (n - 1)(n - 2)(n - 3)....(b)...(a)....(1),

Then (n - 1)! is a multiple of ab and so ab ( = n) divides (n - 1!)

Case ②

n = c² , where c is composite, c < n

but since c is composite, then c can be rewritten as the product of prime factors.

Which means c² can be rewritten into the form ab → back to case ①


Case ③

n = d², where d is prime and d < n

(n - 1)!

= (n - 1)(n - 2)...(n - d)...(d)...(1)

(Since d < n means n - d > 0, so n - d is one of the factors of (n - 1)!)

= (d² - 1)(d² - 2)...(d² - d)...(d)...(1)

= (d² - 1)(d² - 2)...(d)(d - 1)...(d)...(1)

= (d² - 1)(d² - 2)...(d²)(d - 1)...(1)


So (n - 1)! is always a multiple of d², which means that n divides (n - 1)!


HOWEVER,

This is provided that d² - d > d.

Eg. If d = 3, then 3² - 3 = 6 > 3


d² - d > d
d² - 2d > 0
d(d - 2) > 0

d > 0 or d > 2 → overall d > 2
(We already know that d cannot be 1 so no need to consider)


So the exception is when d = 2

n - d = d² - d = 2² - 2 = 4 - 2 = 2

This means that there will only be 1 factor d and not two distinct factors d and d² - d)


i.e

When d = 2, n = 2² = 4

(n - 1)! = 3 x 2 x 1

There is only one '2' in the product. So not divisible by 2²



But when d = 3, n = 9

(n - 1)!
= 8!
= 8 × 7 × (3² - 3) × ... × 3 × 2 × 1
= 8 × 7 × (3 × 2) × ... 3 × 2 × 1

There are two '3's so definitely divisible by 9

Thanks man rly appreciate it

No problem. Hope it's understandable.