Others (Thanks, Larry and Charlie!) have already identified that n = 1 and n = 2 are solutions.
Suppose n is >= 4 and even
Then n! Is a multiple of 8 and so is the first term in the expression. The second expression can be written as 33 * 169 ^ (n/2) which is also equal to 1 mod 8 since both 33 and 169 = 1 mod 8. So the overall expression is equal to 5 mod 8. But all perfect squares are equal to 0, 1, or 4 mod 8, so if n >= 4 and even, there are no solutions.
Now suppose n is >= 7 and odd
Then n! Is a multiple of 7 and so is the first term in the expression. Now, 33 = -2 mod 7 and 13 = -1 mod 7, so the product is 2 mod 7 when n is odd. And the overall expression is then 6 mod 7. But all perfect squares are either 0, 1, 2, or 4 mod 7 so there are no odd solutions with n >= 7.
That leaves only the cases of n = 3 and n = 5 unaccounted for. For n = 3, the expression’s value (72733) is prime and so not a perfect square, and for n = 5 the expression factors as 59*61*3407, also not a perfect square.
So n = 1, n = 2 are the only solutions.
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Posted by Paul
on 2021-06-18 15:31:26 |
proof
