Prove: If x-y is even, then 15|N where N = xy(x+y)(x-y)(x^2 + y^2)?

Showing 3 divides N is easy so let me do 5 first.

x = y + 2k, so x^2 + y^2 = 2y^2 + 4ky + 4k^2 = 2(y^2 + 2ky + 2k^2) = 2((y+k)^2 + k^2). If k = 0 then x = y, and x-y is divisible by 5, so let's assume k <>0. Then x^2 + y^2 = 0 mod 5 ==> (y+k)^2 + k^2 = 0 mod 5.

The only nonzero squares mod 5 are 1 and 4. If k^2 = 1 mod 5 and (y+k)^2 = 4 mod 5 (or vise-verse), then x^2 + y^2 = 0 mod 5, and you are done. If (y+k)^2 = k^2 mod 5 then y+k = +/-k mod 5. If y + k = k mod 5 then y = 0 mod 5, and you are done. If y + k = -k mod 5 then x = y+2k = 0 mod 5 and you are one. This proves that xy(x^2+y^2) is always divisible by 5 if x - y is a nonzero even integer, and that N is always divisible by 5 even when k = 0.

Divisibility by 3 is as follows. Note (x+y)(x-y) = x^2 - y^2, and the only nonzero square mod 3 is 1. So if 3 doesn't divide x or y then both x^2 and y^2 = 1 mod 3,, so x^2 - y^2 = 0 mod 3. This shows that 3 always divides xy(x+y)(x-y), and so also divides N.

Note also that if y is odd then so is x. If x = 2r + 1 and y = 2s+1 then x^2 = 4r(r+1) + 1 an y^2 = 4s(s+1) + 1, which means that x^2 + y^2 is divisible by 2, and x^2 - y^2 is divisible by 8 (because r(r+1) is even). So N is divisible by 16. If y is even then so is x and xy is divisible by 4, x^2 + y^2 is divisible by 4 and x^2 - y^2 is divisible by 4. So N is always divisible by 16.

This all shows that N is divisible by 240 = 15*16.

N = xy(x + y)(x – y)(x² + y²)

In modulo 3

If x = 0 or y = 0 then xy = 0 so N = 0 and 3|N

If x = 1 and y = 1 then x – y = 0 so N = 0 and 3|N

If x = 1 and y = 2 then x + y = 0 so N = 0 and 3|N

If x = 2 and y = 1 then x + y = 0 so N = 0 and 3|N

If x = 2 and y = 2 then x – y = 0 so N = 0 and 3|N

In modulo 5

If x = 0 or y = 0 then xy = 0 so N = 0 and 5|N

If x = 1 and y = 1 then x – y = 0 so N = 0 and 5|N

If x = 1 and y = 2 then x² + y² = 0 so N = 0 and 5|N

If x = 1 and y = 3 then x² + y² = 0 so N = 0 and 5|N

If x = 1 and y = 4 then x + y = 0 so N = 0 and 5|N

If x = 2 and y = 1 then x² + y² = 0 so N = 0 and 5|N

If x = 2 and y = 2 then x – y = 0 so N = 0 and 5|N

If x = 2 and y = 3 then x + y = 0 so N = 0 and 5|N

If x = 2 and y = 4 then x² + y² = 0 so N = 0 and 5|N

If x = 3 and y = 1 then x² + y² = 0 so N = 0 and 5|N

If x = 3 and y = 2 then x + y = 0 so N = 0 and 5|N

If x = 3 and y = 3 then x – y = 0 so N = 0 and 5|N

If x = 3 and y = 4 then x² + y² = 0 so N = 0 and 5|N

If x = 4 and y = 1 then x + y = 0 so N = 0 and 5|N

If x = 4 and y = 2 then x² + y² = 0 so N = 0 and 5|N

If x = 4 and y = 3 then x² + y² = 0 so N = 0 and 5|N

If x = 4 and y = 4 then x – y = 0 so N = 0 and 5|N

By symmetry x and y can be swapped and still 5|N

So we don't need to know that x - y is even because it is always true that if x and y are integers then 15|N where N = xy(x + y)(x – y)(x² + y²)

Let us prove some preliminary results first:

(a) For two integers x and y, if 3∤x and 3∤y, then 3 divides either x - y or x + y.

Proof: There are integers a,b,c and d such that x = 3a + b and y = 3c + d (here b and d are either 1 or 2). so that x - y = 3(a - c) + (b - d) and x + y = 3(a + c) + (b + d). If b = d, then x - y = 3(a - c), which is clearly divisible by 3. If b and d are different, then b + d = 3 so that x + y = 3(a + c + 1), which is also divisible by 3.

(b) If 3 divides either x or y, then one of the two numbers has remainder 0 when divided by 3 and the other has remainder 2. As such, you would have that xy is divided by three.

This proves for us that one of the terms in the product must be divisible by three! Let's show that one is divisible by 5, and we will build our proof thereafter.

(c) Suppose for integers x and y, x - y is even. Then 5|x, 5|y, or 5|(x² + y²).

Proof: If this occurs, then either both remainders when divided by 5 are 0, which means that 5 divides both x and y. One remainder is 0 and the other is 2, implying 5 divides either x or y. One remainder is 1 and the other is 3, implying the 5|(x² + y²) since the sum of the squares of the remainders would be 10. One is 2 and the other is 4, which implies that 5|(x² + y²) since the sum of the squares of the remainders would be 20.

With these three remarks, you can see that exactly one term in N is divisible by 3 and another is divisible by 5. Thus, 15|N.

Evidently, x, y are both even or both odd. If either 3|x or 3|y then 3|N. If 3 divides neither x nor y, then it leaves a remainder 1 or 2 when it is divided into either. The possibilities are (1) x≡1 (mod3) & y≡1 (mod3), (2) x≡1 (mod3) & y≡2 (mod3), (3)x≡2 (mod3) &y≡2 (mod3) and (4) x≡2 (mod3) &y≡1 (mod3). Case(1): x²≡1 (mod3), y²≡1 (mod3)→(x-y)(x+y)= x²- y²≡0(mod3)→3|(x-y)(x+y). Case (2): x²≡1 (mod3), y²≡4 (mod3) ≡1 (mod3) → x²- y² ≡0 (mod3) →3|(x²- y²). Case (3): x²≡4 (mod3), y²≡4 (mod3) →3|(x²- y²), as before. Case (4): similar to case (2) and 3|(x²- y²).

We next have to show that 5|N. If either 5|x or 5|y, then 5|N. If neither of these hold, then on division by 5 the remainder is 1, 2, 3, or 4. Indeed we can show quite easily that x², y²≡1 (mod5) or ≡-1(mod5). [I am going to skip the steps in this demonstration and assume that you can do it EASILY for yourself]. You can also easily see that this will either lead to 5|( x²- y²) when x²≡1 (mod5) & y²≡1 (mod5) or x²≡-1 (mod5) & y²≡-1 (mod5). And to 5|( x²+ y²) when

x²≡1 (mod5) & y²≡-1 (mod5) or vice versa.

Edit: if you are really stuck in filling in my gaps, do post and I'll do so.

Edit: I’ve just thought another approach to the divisibility by 5 case.

We have x^(5-1) ≡1 (mod5) by Fermat’s Little theorem→x⁴≡1 (mod5), y⁴≡1 (mod5)→x⁴ - y⁴≡0 (mod5) →5|(x⁴ - y⁴)→5|N.

a difficulty arises once you pass from ( 2 – 3 )^2 = ( 4 – 3 )^2 to 2 - 3 = 4 - 3 because of the fact the expression on the left is adverse mutually as the expression on the properly suited is beneficial (an untrue fact) once you're taking a sq. root, the result's + or - the basis. +-(2 - 3) = +-(4 - 3) and basically p.c.. the mixtures that are real, like +(2 - 3) = -(4 - 3) or -(2 - 3) = (4 - 3) including 3 to the two factors of the two of those equations supplies yet another real fact.