Determine all possible values of a prime number P such that P
3-4P+9 is a perfect square.
Note: Computer program aided solutions are welcome, but a semi-analytic solution is preferred.
There are only 3: {2, 7, 11}
f(2) = 9 = 3^2
f(7) = 324 = 18^2
f(11) = 1296 = 36^2
Note the function of p is always 0 mod 3. But if p=3, f(p) = 24 and is not prime. Going further, since the expression is a square there must be another factor of 3; so it must be 0 mod 9.
Further, if we divide it by 9, the result will also be a square:
f(p)/9 = p * (p^2-4)/9 + 1
Since p mod 9 cannot be 0 since it is prime, (p^2-4) is 0 mod 9.
That happens when p is 2 mod 9 or 7 mod 9.
Plenty of primes satisfy that condition, the first several are:
p mod9
2 2
7 7
11 2
29 2
43 7
47 2
61 7
79 7
83 2
97 7
Consider again, the function divided by 9 for the first several primes that are either 2 or 7 mod 9.
2: 2 * 0/9 + 1 = 2 * 0 + 1 = 1 a square
7: 7 * 45/9 + 1 = 7 * 5 + 1 = 36 a square
11: 11 * 117/9 + 1 = 11*13 + 1 = 144 a square
29: 29 * 837/9 + 1 = 29*93 + 1 = 2698 not
43: 43 * 1845/9 + 1 = 43*205 + 1 = 8816 not
47: 47 * 2205/9 + 1 = 47*245 + 1 = 11516 not
2 turns out to be a special case of 0 + 1 being 1
7 and 11 are a different kind of special case such that the formula resolves to something of the form (a-1)*(a+1) + 1 = a^2 - 1 + 1 = a^2
I haven't proved that the above pattern can never be (a-1)*(a+1) + 1 for p > 11, but looks like a good bet.
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Posted by Larry
on 2022-12-12 10:14:19 |
Almost a proof
