Prove that pic related is always a multiple of 10 for any integers a and b. Thought some of you might enjoy this problem

algebraic preferably

x^p = x mod p (Fermat's little theorem)
x^2 = x mod 2
(a)*(b^5) = a*b = (a^5)*(b) mod 2
(a)*(b^5) - (a^5)*(b) = 0 mod 2

x^5 = x mod 5
(a)*(b^5) = a*b = (a^5)*(b) mod 5
(a)*(b^5) - (a^5)*(b) = 0 mod 5

Reduce it all mod 10. The only squares mod 10 are 1, 4, 6, 9, so (b^2 - a^2)*(b^2 + a^2) will always be divisible by either 10 or 5, and in the case it's divisible by 5, then ab will be even.

fucking beautiful man. I thought up a solution using the ab(a-b)(a+b)(a^2+b^2) expression but yours is just so fucking simple and elegant.

How is that an answer? pardon me for being a dumbass but I don't see how this proves the divisibility.

he's shown that [math]5|a\cdot b^5-a^5\cdot b[/math] and [math]2|a\cdot b^5-a^5\cdot b[/math], and since [math]\gcd(2,5)=1[/math] it follows that [math]10|a\cdot b^5-a^5\cdot b[/math]

nice job btw

n(mod 2)=1 or 0
Theories used to prove this equation
a+b= (a(mod x)+b(mod x))
ab(mod x)=(a(mod x)*b(mod x)) (mod x)
if a=1 mod 2 then a^5=1, if a^5 =0 mod 2 then a^5=a mod 2 [from the fact that 1^5=1 and 0^5 = 0], repeat the proof for b
therefore a= a^5 mod 2, b^5=b mod 2
Therefore we know that both a^5b and b^5a have the same modality, as such went they are subtracted they cancel out and we are left with a zero which mean that such number will be divisible by the modular base, in this case 2
As we know that 10=2*5, than every number that is divisible by both 2 and 5 will be a multiple of 10.
use fermat's little theorem to prove that such a statement will always be divisible by 5 and were done.

Sort of brute force, but it works:
ab(b^4 - a^4)
If a or b is even, then 2|ab(b^4 - a^4)
Say both a and b are odd. Then b^4 and a^4 are odd, and b^4 - a^4 is even. So 2|ab(b^4 - a^4).

We have shown that ab(b^4 - a^4) divides 2 in all cases. How about 5?

If a or b is a multiple of 5, we are done, and 10|ab(b^4 - a^4).

If neither a nor b is a multiple of 5, consider b^4 - a^4 = (b^2 + a^2)(b + a)(b - a)

If (b + a) or (b - a) is a multiple of 5, again we are done. Assume neither a, b, b+a, or b-a is a multiple of 5.

We know that a = 1, 2, 3, or 4 mod 5, and b = 1, 2, 3, or 4 mod 5. Consider all 16 pairs (a,b)
in {1,2,3,4}x{1,2,3,4}.

Firstly, if a=b mod 5, then 5|b-a. But we have assumed this not to be true, therefore, a != b mod 5.

Also, our elements (a,b) cannot add to a multiple of 5 by assumption, so we cannot pick (1,4),(4,1),(2,3), or (3,2).

As a^2 + b^2 is symmetric, WLOG take a < b. Then we only have to check (1,2),(1,3),(2,4),(3,4).

1^2 + 2^2 = 5 is a multiple of 5
1^2 + 3^2 = 10 is a multiple of 5
2^2 + 4^2 = 20 is a multiple of 5
3^2 + 4^2 = 25 is a multiple of 5

Therefore, a^2 + b^2 is a multiple of 5 whenever a, b, b+a, and b-a are not, but the original quantity is even, so we have that ab(b^4-a^4) = ab^5 - a^5 b is a multiple of 10.

>mfw i will never be as clever as this
WHAT"S THE FUCKING POINT AHHHHHHHHHHHHHHHHHHHHH

Now that's fucking neato. How the hell did you get the idea for using Fermat?

Not that user but if a problem involves prime exponents and divisibility, Fermat's Little Theorem is always a good place to start.

in what sort of class do you get to prove comfy equations like this a lot?

intro to proofs

I haven't done any number theory in a few years, so this solution's is a bit messy.
[math]ab^{5} - a^{5}b //
= ab(b^{4} - a^{4}) //
=ab(b^{2} - a^{2})(b^{2} + a^{2}) //
= ab(b + a)(b - a)(b^{2} + a^{2}) [/math]
Now if a or b is even, then this expression is even. If a and b are odd, then it is clear that a + b is even. Hence this expression is even for all integers a, b.
Now we need to show that this expression is always a multiple of 5.
First we assume that a and b are not divisible by 5 (or else we are done).
So let a = 5i + p and b = 5j + q, for some integers i, j, p, q where 0 < p, q < 5.
So take the modulus of the expression, to obtain
p * q * (p + q) * (p - q) * (p^2 + q^2)
So for it to be divisible by 5, we just need to show that this is a multiple of 5 whenever p, q are in {1,2,3,4}
There are 10 possible cases here (p=1,q=1, p=1,q=2, ..., p=4,q=4)
I then wrote a simple function (in the attached picture) in Python to check each solution, which confirmed that the expression is divisible by 5 in all cases.
As the expression is divisible by both 2 and 5, it is divisible by 10 and we are done.

Whoops, I thought // was newline in Latex, but we all make mistakes.

I did a lot of this stuff in my first year Number Theory course.

this is the first decent Mathematics thread
I've seen here in 2018, congratos lads