If we start with the identity n=((n+1)/2)²-((n-1)/2)², then, if n is an odd whole number, it can clearly be written as a difference of squares (of whole numbers). If n is the square of an odd number, i.e. n=(2k+1)², then we have the identity (2k+1)²= (2k²+2k+1)²-(2k²+2k)². Hence every odd number, other than 1, must appear in some Pythagorean triple. Also, every even number, other than 2, appears in some Pythagorean triple.

Cubes can also be represented as differences of squares, e.g. n³ = (n(n+1)/2)² – (n(n-1)/2)².

An application of the above identity is to∑_1≤k≤nk³ = (n(n+1)/2)².

For higher powers, if we try n^a = (n^b(n^c + 1)/2)² – (n^b(n^c-1)/2)² ,

where a, b and c are whole numbers, then a=2b+c. So, if a=5, we can have b=2 and c=1 or b=1 and c=3.

Hence, for n⁵(say) we can have either n⁵ = (n²(n+1)/2)² – (n²(n-1)/2)² or

n⁵= (n(n³+1)/2)²– (n(n³-1)/2)².

Divisibility by 3

If (x,y,z) is a PPT (Primitive Pythagorean Triple) then z² = x² + y². Looking at examples of PPT’s it is evident that exactly one of x and y is divisible by 3. This follows from an application of n³≡ n mod 3 for any integer n.

Proof

If z²= x²+ y² then 2x²y² = z⁴ – x⁴ – y⁴ ≡ z²– x²– y² mod 3≡ 0 mod 3 and so 3|2x²y². Hence, 3 divides x or y or both. 3 can’t divide both x and y because then 3 would then divide z, since z²= x²+ y², meaning that (x,y,z) is not primitive. Hence the result.

For positive integer powers of positive integers it is clear that these can be expressed as differences of squares of positive integers. What is also clear is that in all cases at least one of these squares must be divisible by 3.

Case 1 Odd powers, 3 and higher

n^2k+1 = (n^k(n+1)/2)² – (n^k(n-1)/2)² , k≥1

Case 2 Even powers, 4 and higher

n^2k = (n^k-1(n² + 1)/2)² – (n^k-1(n² – 1)/2)² , k≥2

Obviously if 3|n then both squares in the above cases are divisible by 3. Otherwise, in Case 1 either the first or the second square must be divisible by 3 depending on n, while in Case 2 the second square is always divisible by 3.

Divisibility by 5

If (x,y,z) is a PPT (Primitive Pythagorean Triple) then z²= x²+ y². Looking at examples of PPT’s it is evident that exactly one of x, y and z is divisible by 5. This follows from an application of n⁵≡ n mod 5 for any integer n.

Proof

If z²= x²+ y² then z⁶ = x⁶ + 3x⁴y² + 3x²y⁴ + y⁶ = x⁶ + y⁶ + 3x²y²z².

Hence, 3x²y²z² = z⁶ – x⁶ – y⁶≡ z² – x² – y² mod 5≡ 0 mod 5. Therefore 5|3x²y²z² and so the result follows.

If we look at n^2k= (n^k-1(n²+ 1)/2)²– (n^k-1(n²– 1)/2)², k≥2, then the product of the terms to be squared gives n^2k-2(n⁴ – 1)/4 = n^2k-3n(n⁴– 1)/4≡ 0 mod 5, so at least one of the squared terms must be divisible by 5.