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Algebraic Manipulation
Darealzolt   1
N Apr 30, 2025 by Soupboy0
Find the number of pairs of real numbers $a, b, c$ that satisfy the equation $a^4 + b^4 + c^4 + 1 = 4abc$.
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Darealzolt
Apr 30, 2025
Soupboy0
Apr 30, 2025
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Darealzolt
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#1 h  Apr 30, 2025, 1:25 PM
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Find the number of pairs of real numbers $a, b, c$ that satisfy the equation $a^4 + b^4 + c^4 + 1 = 4abc$.
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Soupboy0
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#2 h  Apr 30, 2025, 2:36 PM
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$(a, b, c) = (1, 1, 1), (-1, -1, 1)$ and permutations. Note that if $a, b, c > 0$, then $a^4+b^4+c^4+1 \ge 4abc$ by AM-GM, with equality at $a=b=c=1$, which is the only solution for $a, b, c > 0$. Also note that if $1$ or all of $a, b, c$ are negative, then there are no solutions by the Trivial inequality. Also note that when $2$ of $a, b, c$ are negative, then this case is symmetric to $a, b, c > 0$. Therefore, there are $4$ solutions
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