Problems & Puzzles: Puzzles

 Puzzle 452. (a^2)!+2^(a^2) = b*(1+a^2) Juan López sent the following puzzle: Are there infinitely many solutions of the diophantine equation (a^2)!+2^(a^2)=b*(1+a^2) in positive integers? Observe that the odd primes of the form 1+a^2 satisfy the equation by application of Wilson Theorem and Fermat's Little Theorem.

Contributions came from Luke Pebody, Rustem Aidagulov & farideh Firoozbakht.

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Luke wrote:

Setting x=a^2, the condition in puzzle 452 is (x+1)|x!+2^x.

This is easy to solve.
By Wilson and Fermat, this is true for x+1 an odd prime.
It is not true for x+1=2: 2 is not a factor of 1+2=3.
It is not true for x+1=4: 4 is a factor of 6+8=14

Now suppose x is any solution where x+1 is not an odd prime or in the set {2,4}.
Then (x+1) is a factor of x!. Thus the condition is equivalent to x+1|2^x.

Then (x+1) must be a power of 2. Suppose x=2^t-1.
Then the condition can be restated as 2^t|2^(2^t-1), or t<=2^t-1.
This is clearly true for all non-negative t (by induction if necessary).

Thus the solutions to (x+1)|x!+2^x are:
x=p-1 for odd primes p
x=2^t-1 for t=0 or t>=3.
==============

If t>=3 and x=2^t-1 then x is an odd number which is not equivalent to
1 (mod 8), and thus is not a square. Thus the solutions to Puzzle 452
are a=0 and those numbers for which a^2+1 is prime.

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Rustem wrote:

If n is not prime, then n|(n-1)!, therefore if 1+a^2 is not prime, then 1+a^2 =2^k or 1+i.a = +/-(1+/-i)^k. It give only one solution a=1, but 1!+2 not divisible1+1^2=2 . Therefore question equivalent to known conjecture: there are infinitely many primes form 1+x^2.

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Farideh wrote:

Dear Carlos,

As he pointed out numbers x such that x^2+1 is an odd prime are solutions of the Diophantine equation (x^2)!+2^(x^2) = 0 (mod x^2+1) (*).
We can easily show that there is no other solution.

Proof:

Case 1. If x^2+1=2 (an even prime) then x=1 and 2 doesn't divide 1!+2^1.

Case 2: If x^2+1 is a composite number then it's obvious that x^2+1 divides (x^2)!,
but if x^2+1 divides 2^(x^2) then x^2+1=2^k for a number k>1 so x^2=3 (mod 4) which
is impossible. Hence in this case x^2+1 doesn't divide (x^2)!+2^(x^2).

So in both cases x can"t be a solution for (*) and the proof is complete.

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Hence the question of this puzzle is equivalent to the following known conjecture.

Conjecture: There are infinitely many primes of the form n^2+1.

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