Question

x^3+y^3=10000. Verify that the above equation has no integer solution where x doesn't equal to y​

Answer 1

Well, first of all, 7∤100007∤10000 -- in particular, 10000≡4(mod7)10000≡4(mod7) . If we looked at (mod5)(mod5), it would be useless because we can have x,yx,y be multiples of 55 and our method would not work.

Second of all, 7=23−17=23−1. Thus, we have (2n)3≡n3(mod7)(2n)3≡n3(mod7) and for n∈{1,2,3}n∈{1,2,3} the residue is either −1−1 or 11. Thus, we can also say for odd numbers that (7−2n)3≡(−2n)3≡−n3(mod7)(7−2n)3≡(−2n)3≡−n3(mod7), so that's an explanation of why the solution chose 77 (it can only take on residues of −1,0,1(mod7)−1,0,1(mod7))

Answer 2

Ҩմҽʂէìօղ-

->> x^3+y^3=10000. Verify that the above equation has no integer solution where x doesn't equal to y

ᗅᘉSᙎᙍᖇ-

An integer cube will always leave a remainder of either 0,1,60,1,6 when divided by 77 (I suppose that I will leave this to you to prove), while 1000010000 leaves a remainder of 44 when divided by 77.

So we see that if we divide x3+1000x3+1000 by 77, it must leave a remainder of either 4,5,34,5,3 (by summing up both of their remainders in the ring of integers modulo 77).

On the other hand, y3y3 is an integral cube. As disused earlier, all integer cubes must leave a remainder of either 0,1,60,1,6 when divided by 77.

Since the remainder of both sides when divided by 77 will never be equal, there are no integer solutions.

A possible alternative would be to rearrange the given problem into:

y3−x3=10000

y3−x3=10000

Now, we factorize this to get:

(y−x)(y2+xy+x2)=10000

(y−x)(y2+xy+x2)=10000

77 is a good number to try because the multiplicative group of non-zero residues modulo a prime pp is cyclic of order p−1p−1. [Order p−1p−1 is very easy, cyclic takes a bit more work, note 7−1=67−1=6 is divisible by 33]. This means that the cubes modulo 77 will be 00 and 7−13=27−13=2 other numbers which turn out to be ±1±1. This is a particularly simple set of things to be working with, and it is easy to see that

{0,±1}+4={3,4,5}

{0,±1}+4={3,4,5}

If this hadn't worked, the primes to try are the ones of the form 6n+16n+1 - the cubes mod 1313 are 0,±1,±80,±1,±8 - there are 13−13=413−13=4 non-zero cubes modulo 1313. 1000010000 leaves remainder 33.

{0,±1,±8}+3={3,2,4,11,−5}

{0,±1,±8}+3={3,2,4,11,−5}

but modulo 33 we have −5=8−5=8 and we have more work to do. As numbers increase, we have more cases to consider.

If we choose a prime p=6n−1p=6n−1 we find that every element is a cube (the cyclic group of order 6n−26n−2 has no elements of order 33, which would cube to 11), and this approach gets us nowhere.