MathLabs

Problem 4

Determine all pairs (x,y)(x,y) of integers such that 1+2x+22x+1=y21+2^{x}+2^{2x+1}=y^{2}.
Step 2 of 7: Solve the zero-exponent case
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x=0⟹y2=4⟹(x,y)=(0,±2)x=0\Longrightarrow y^2=4\Longrightarrow (x,y)=(0,\pm2)
Detailed analysis

For x=0x=0, the equation gives 1+1+2=41+1+2=4, so y=±2y=\pm2. This yields (x,y)=(0,±2)(x,y)=(0,\pm2).