Three equal evaluations determine quadratic coefficients

Let Bob's quadratic be $ax^2+bx+c$ and Alice's be $dx^2+ex+f$ . We know that,

$a+b+c=d+e+f$ ,

$4a+2b+c=4d+2e+f$ ,

$9a+3b+c=9d+3e+f$ .

Then by subtracting the 2nd equation from the 3rd one we have $5a+b=5d+e$ .

By subtracting the first equation from the second we have $3a+b=3d+3$ .

Now we subtract $3a+b=3d+3$ from $5a+b=5d+e$ to find $2a=2d$ or $a=d$ .

By substituting $a=d$ into our original 3 equations we have,

$a+b+c=a+e+f$ or $b+c=e+f$ ,

$4a+2b+c=4a+2e+f$ or $2b+c=2e+f$ ,

$9a+3b+c=9a+3e+f$ or $3b+c=3e+f$ .

By subtracting the second equation from the first one we obtain $b=e$ so then $c$ must equal $f$ .

Therefore, the quadratics are the same.

Source: InternLM Lean-Workbook, record lean_workbook_12336 (Apache-2.0). The complete source proposition and explicit variable declarations are preserved.