When a = 0
LHS = (x²⁺⁰ + y²⁺⁰)² = (x² + y²)²
RHS = (x² + y²)²⁺⁰ = (x² + y²)²
LHS = RHS

----------

When a > 0

Expand Left hand side (LHS) and Right hand side (RHS) and compare:

LHS = (x²⁺ᴬ + y²⁺ᴬ)² = x²⁽²⁺ᴬ⁾ + 2x²⁺ᴬy²⁺ᴬ + y²⁽²⁺ᴬ⁾
LHS has 3 distinct terms

. . . . . . . . . . . . . . . .2+a
RHS = (x² + y²)²⁺ᴬ = ∑ C(2+a,k) x²⁺ᴬ⁻ᴷ yᴷ, where C(2+a,k) = 2+a choose k
. . . . . . . . . . . . . . . .k=0

RHS has 2+a+1 = (3+a) distinct terms (3+a > 3, since a>1)

LHS ≠ RHS (since they have different number of distinct terms)

----------

For a < 0 → check a = -1, a = -2, a < -2

When a = -1
LHS = (x²⁻¹ + y²⁻¹)² = (x + y)² = x² + 2xy + y²
RHS = (x² + y²)²⁻¹ = x² + y²
LHS ≠ RHS

When a = -2
LHS = (x²⁻² + y²⁻²)² = (x⁰ + y⁰)² = (1+1)² = 4
RHS = (x² + y²)²⁻² = (x² + y²)⁰ = 1
LHS ≠ RHS

----------

When a < -2

In this case we get into negative exponents.
I haven't quite figured out how to prove this algebraically yet.

Permalink | Report