MHB Possible webpage title: Which Primes Make p!+p a Perfect Square?

  • Thread starter Thread starter alexmahone
  • Start date Start date
  • Tags Tags
    Square
AI Thread Summary
The discussion focuses on identifying prime numbers p for which the expression p! + p is a perfect square. The initial findings confirm that for p = 2 and p = 3, the expression yields perfect squares. Using Wilson's theorem, the relationship p! + p = p[(p-1)! + 1] is established, indicating that (p-1)! + 1 must be divisible by p. Further analysis suggests that for larger primes, the factorial grows too rapidly, making it unlikely for p! + p to equal a perfect square. The conversation highlights the complexity of the problem and the difficulty in extending the findings beyond small primes.
alexmahone
Messages
303
Reaction score
0
Find all prime numbers p for which p!+p is a perfect square.

My thoughts: 2!+2 and 3!+3 are perfect squares.
p!+p=p[(p-1)!+1]
By Wilson's theorem, (p-1)!+1 is divisible by p. Now I'm stuck.
 
Last edited:
Mathematics news on Phys.org
Alexmahone said:
Find all prime numbers p for which p!+p is a perfect square. My thoughts: 2!+2 and 3!+3 are perfect squares. p!+p=p[(p-1)!+1] By Wilson's theorem, (p-1)!+1 is divisible by p. Now I'm stuck.
Well, you are almost done. (p - 1)! + 1 = kp p! + p = kp^2 so we see that k must be a perfect square to satisfy your condition. let k = m^2 p!+p = (mp)^2 also p!+p = c^2 c^2 = (mp)^2 p = c/m now if m = 1 then c = p and p!+p = p^2 that's only going to happen for 2!, 3! cause they have few products. As the primes get larger than 3 the p! has too many products destroying any possibility that p!+p = p^2 Now I'm stuck cause i can't think of a good reason why m can't be any other positive integer...

---------- Post added at 07:05 PM ---------- Previous post was at 07:03 PM ----------

why are all the sentences getting bunched up??
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
View full post »

Similar threads

I Questions regarding Kurepa's Conjecture
Replies
1
Views
2K
MHB Finding Perfect Square Combinations: $m,n\in N$
Replies
3
Views
1K
Proof: Twin Primes Always Result in Perfect Squares
Replies
3
Views
2K
The only prime p for which 3p+1 is a perfect square is p=5?
Replies
15
Views
4K
MHB Which Prime Numbers Make p^2 + 1007 Have Less Than 7 Divisors?
Replies
2
Views
2K
Show that the sum of twin primes ## p ## and ## p+2 ## is divisible?
Replies
17
Views
3K
MHB Max 3-Digit Prime Divisor of 2000!/(1000!1000!)
Replies
1
Views
1K
MHB Seeking Help with a Calculation
Replies
7
Views
3K
MHB Dimensional regularizatoin of an integral
Replies
7
Views
2K
If ## p ## and ## p^{2}+8 ## are both prime numbers, prove that....
Replies
16
Views
3K

Hot Threads

Recent Insights

Back
Top