Math Forum / Mathematics / Research / October 2008

Let p be any prime between n and n / 2 (but exceeding n / 2).
Then p^2 divides n^2, p divides n-factorial, p^2 doesn't
divide n-factorial, so p divides left side but p^2 doesn't,
so left side is not a square.
--
Gerry Myerson

I think I can oblige.

n^2! + n! = (n^2.(n^2-1).(n^2-2)....(n+1)+1)n! Just multiplying out.

Consider n^2.(n^2-1).(n^2-2)....(n+1)+1

It cannot have a factor between n+1 and n^2 for if we divide by such a
number we always obtain a remainder of 1. This number must either be
prime or the product of large primes. If crucially we take the numbers
between n+1 and n^2 that are divisible by n we find that every value
is there.

Hence n! and n^2.(n^2-1).(n^2-2)....(n+1)+1 cannot have a common
factor. Hence n! must be a square. If we can say that there is a prime
number between –n and n we know that m^2 cannot exist. For this prime
occurs once and once only in n^2 + n! for it occurs once and once only
in n! A prime >–n cannot be a factor of any number <=n except itself.

If there is always a prime between –n and n your statement is thus
proven. My question is Does prime number theory prove this FOR
CERTAIN? This statement is certainly true. You can see this on the
basis of the statistics of the distribution of primes. I can guarantee
therefore that you will never find

m^2 = n^2! + n!

but it would be nice to have absolute rigor on prime number
distributions.

 - Ian Parker

Hi,

It appears that some responders misunderstood my conjecture ...
Let me reiterate my conjecture once more:

The sum

n! + n^2

for n >=1

doesn't yield the perfect square

I also was advised for the additional clarity to present above in the
TEX format as:

$n! + n^2 \ne m^2$

Also any one could check the OEIS's A004664 to see what is involved
there.

Thanks,
Best Regards,
Alexander R. Povolotsky
=========================================================

FYI - Tony Noe came up with the partial proof for my conjecture (when
"n" is prime)
- see below.

Tony Noe also extended the calculations up to n = 10^5
(my original calculation was up to n = 30,000) and found
that my conjecture holds to be TRUE for entire range.

Thanks,
Best Regards,
Alexander R. Povolotsky

I complement my conjecture, which I originally posted in this thread,
namely: n! + n^2 != m^2       for n>=1, m>= 0

with another conjecture

n! + Sum(j^2, j=1, j=n) != m^2       for n>=1, m>= 0

I checked using PARI program
isSquare(n)={local(a);a=sqrt(1/6*n*(n+1)*(2*n
+1)+n!);if(floor(a)==ceil(a),print1("n="n"\n"))}
for(n=1,30000,isSquare(n))
that indeed
n! + Sum(j^2, j=1, j=n)
doesn't yield perfect square up to n=30,000

Thanks,
Best Regards,
Alexander R. Povolotsky