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  "body": "\n## Example\n\nWrite $A = 1 + x$, $B = 1 + x + x^2$, $C = 1 + x^3$ and $D = 1 + x^2 + x^4$, where $x^2$ means $x \\cdot x$ and so on.\n*Wilkie's identity* is\n$$(A^y + B^y)^x \\cdot (C^x + D^x)^y = (A^x + B^x)^y \\cdot (C^y + D^y)^x.$$\nIt is an [[identity|PKD440G2XW]] in the signature $1, +, \\cdot$, exponentiation.\n\nCheck that it holds for all positive integers $x, y$.\nLet $Q = 1 - x + x^2$, a positive integer for $x \\ge 1$.\nThen $C = A \\cdot Q$ and $D = B \\cdot Q$.\nSo $C^x + D^x = Q^x (A^x + B^x)$ and $C^y + D^y = Q^y (A^y + B^y)$.\nBoth sides become $(A^y + B^y)^x \\cdot (A^x + B^x)^y \\cdot Q^{xy}$.\n\nAlex Wilkie showed that this identity cannot be derived from the eleven high school identities, which answers [[Tarski's high school algebra problem|RCRD8CAVB4]] negatively.\nThe check above uses the term $1 - x + x^2$, which needs subtraction and so is not a term of the signature.\n\n## Explanations\n\n### L1 human\n\nBoth sides of Wilkie's identity are built only from $1$, addition, multiplication and powers, and they are always equal for positive whole numbers.\nThe quick proof factors $1 + x^3$ and $1 + x^2 + x^4$ with the help of $1 - x + x^2$, which uses a minus sign.\nSchool rules have no minus sign, and Wilkie proved that they cannot prove the identity in any other way.\n",
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