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Add to Factorial, Get Perfect Power (Posted on 2025-02-21)

Find all possible pairs (n, k) of positive integers that satisfy this equation:

n! + n = n^k

No Solution Yet Submitted by K Sengupta

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computer findings

| Comment 1 of 2

for n=vpa(1):2000

lhs=factorial(n)+n;

logrhs=log(lhs);

k=logrhs/log(n);

if abs(k-round(k))<.00001

disp([n,factorial(n),k])

end

finds


  n    n!    k
[2.0, 2.0, 2.0]
[3.0, 6.0, 2.0]
[5.0, 120.0, 3.0]

Posted by Charlie on 2025-02-21 10:29:29

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