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  "aliases": [
    "binary-exponentiation",
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  "body": "\n## Summary\n\nModular exponentiation computes $b^e \\bmod m$ for integers $b$, $e \\ge 0$ and $m \\ge 1$.\nThe binary method writes $e$ in base $2$. It squares $b$ again and again modulo $m$, and it multiplies the squares $b^{2^i}$ for the bits $i$ of $e$ that are $1$.\nIt needs about $2 \\log_2 e$ multiplications modulo $m$, and each intermediate number is smaller than $m^2$.\n\n## Remarks\n\nThe method is correct because of [[compatibility with multiplication|072D5G4SK6]]: one can reduce modulo $m$ after each multiplication.\nThe [[Fermat primality test|0N5ZSZMYH6]], the [[RSA cryptosystem|0HM584C465]] and the [[Diffie–Hellman key exchange|01TKKESHSB]] all use modular exponentiation.\nThe inverse problem, to find $e$ from $b$, $m$ and $b^e \\bmod m$, is the [[discrete logarithm|09B4X51J09]] problem. It seems much harder.\n\n## Explanations\n\n### L1 human\n\nTo compute $3^{13} \\bmod 7$, write $13 = 8 + 4 + 1$. The squares are $3^1 \\equiv 3$, $3^2 \\equiv 2$, $3^4 \\equiv 4$ and $3^8 \\equiv 2$.\nThen $3^{13} \\equiv 2 \\cdot 4 \\cdot 3 = 24 \\equiv 3 \\pmod 7$.\nYou never need the large number $3^{13} = 1594323$.\n",
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