find integer solution of this equation:x!=y*y?

Question

only the integr and positive

this except x=1,y=1

Answer 1

I think that "FedUp" is definitely going in the right direction. Since a square is teh product of two identical numbers, all prime factors of a square must appear an even number of times.

If x is prime then the factor x only appears once so x! = y^2 cannot be true for any prime x.

Starting with a prime value, call it z, x = z+1 cannot be a solution because (z + 1)! still contains that lone factor of z. In fact, to pair the prime factor of z with another factor of z requires a value of x = 2z because the next integer multiple of z is 2z. If any prime number occurs between z and 2z, it then introduces a new prime factor which must be paired to produce a perfect square. This new prime factor raises the minimum value of x even higher, incorporating new prime factors.

For example 2! = 2 is not a square, it contains a single prime factor of 2. In order to pair this factor you need to go to (2*2)! = 24. This pairs off the 2 but introduces a prime factor of 3. To pair that, you must go to 6!, introducing a factor of 5 requiring 10!. This introduces a factor of 7, requiring 14!

In summary, as long as, for any prime number z1 there exists another prime z2 such that z1 < z2 < 2*z1, there can be no solution to your equation.

The "Prime Number Theorem" deals with the spacing of primes. It states that if the number of primes less than x is called f(x) then f(x) ~ x/ln(x). The spacing (s) between primes z1 and z2 then is:

f(z2) - f(z1) = 1 = z2/ln(z2) - z1/ln(z1)
1 = (z1 + s)/ln(z1 + s) - z1/ln(z1) -> s/ln(z1)
s ~ ln(z1)

The spacing to the next prime is typically proportional to the natural log of the number which is much smaller than the number.

While this is not a proof, since minimum spacing between primes is not assured. It is a pretty fair bet that no solution for x! = y^2 exists past 1.

Answer 2

we want even powers of all primes for x!
pattern for x! for powers of n starting with x=0
x=.... 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ....
n=2 - 0, 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16...
n=3 - 0, 0, 0, 1, 1, 1, 2, 2, 2, 4, 4, 4, 5, 5, 5, 6, 6, 6, 8..
n=5 - 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3..
there is a clear pattern here, and we know it works the same way for all primes. what we want is for a column of numbers in this table to have only even numbers.
well, we see the first solutions at x=0,1, when all the powers are equal to 0. but it seems as if everywhere else, there is bound to be a 1 somewhere.
we find out that other solutions can only occur when there are 2 consecutive primes such that the larger is more than twice as big as the smaller. i'm pretty sure this never happens. could someone confirm this?
however, even if this did occur, (presumably very far down the number line) we would still need to check ALL the smaller primes before to see if they had even powers.

EDIT: ahh... good job, person above me.

Answer 3

i think blighmaster is correct. there will always be a highest prime in the factorial of x and this will not have any counterpart to make it a perfect square. if u can prove this wrong then pls. let us know the solution.

Answer 4

x =0, y=1 or

x=1, y=1

No other solutions are possible.