Prove that infinitely many positive integers n have the property that for any prime p dividing n^{2} + 3, we can find an integer m such that (1) p divides m^{2} + 3, and (2) m^{2} < n.

**Solution**

I started by finding some small n with the required property. This led to: (1) 5^{2} + 3 = 2^{2}7, with 7 | 2^{2} + 3, 2 | 1^{2} + 3; (2) 12^{2} + 1 = 7^{2}3, with 3 | 3^{2} + 3; (3) 23^{2} + 1 = 2^{2}7 19 with 19^{2} | 4^{2} + 3.

That suggested as the first line of attack looking at special n such as 2^{a}7^{b}. That led nowhere.

My second line of attack was to look for relations of the type (a^{2} + 3)(b^{2} + 3) = f(a, b)^{2} + 3. After some playing around, I obtained (m^{2} + 3)( (m+1)^{2} + 3) = M^{2} + 3, where M = (m^{2} + m + 3). We are almost there. Any prime factor of M^{2} + 3 which divides m^{2} + 3 has the required property since m^{2} < M. But that is not true for prime factors of (m+1)^{2} + 3. However, all we have to do is to iterate. We need to split the larger factor:

(m^{2} + 3)( (m+1)^{2} + 3)( (m^{2}+m+2)^{2} + 3) = ( (m^{2}+m+2)^{2} + (m^{2}+m+2) + 3)^{2} + 3.

This shows that any n of the form ( (m^{2}+m+2)^{2} + (m^{2}+m+2) + 3) has the required property.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001