MathLabs

Problem 4

Prove that if nn is a positive integer such that x3−3xy2+y3=nx^3-3xy^2+y^3=n has an integer solution (x,y)(x,y), then it has at least three such solutions. Show that the equation has no integer solutions for n=2891n=2891.
Step 3 of 3: Rule out n=2891n=2891 modulo 77
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In plain words

A small modulus detects a forced common factor, and the cubic degree then amplifies it to an impossible divisibility.

2891≡0(mod7),x3−3xy2+y3≡0(mod7)⟹x≡y≡0(mod7)2891\equiv0\pmod7,\quad x^3-3xy^2+y^3\equiv0\pmod7\Longrightarrow x\equiv y\equiv0\pmod7
Detailed analysis

Since 2891=7⋅4132891=7\cdot413, any solution would satisfy x3−3xy2+y3≡0(mod7)x^3-3xy^2+y^3\equiv0\pmod7. Checking the seven residue classes (as in the source) shows this congruence has only x≡y≡0(mod7)x\equiv y\equiv0\pmod7. Thus 77 divides both xx and yy, so every term of the cubic is divisible by 737^3, forcing 73∣28917^3\mid2891, false.