Let $N = 989 \cdot 1001 \cdot 1007 +320$ and let $p_1 = 989$, $p_2 = 1001$, and $p_3 = 1007$. Claim: $N = (p_1 + 2)(p_2 - 4)(p_3 + 2)$ $N = (p_2 - 4)(p_{1}p_{3} + 2p_1 + 2p_3 + 4)$ $N = p_{1}p_{2}p_{3} - 8p_1 + 4p_2 - 8p_3 +2p_{1}p_{2} -4p_{1}p_{3} + 2p_{2}p_{3}-16$ and $320 = - 8p_1 + 4p_2 - 8p_3 +2p_{1}p_{2} -4p_{1}p_{3} + 2p_{2}p_{3}-16$ Hence $989 \cdot 1001 \cdot 1007 +320 = 991 \cdot 997 \cdot 1009$

Wow... how did you come up with that? Did you use computer calculations? Nice solution anyway.

Well I will admit I calculated N and then looked for some of its prime divisors. I had a hunch that since it had to be proven that it was composite that its primes were somehow related to the the other numbers. So I tried a general form of $N = (p_1+k_1)(p_2+k_2)(p_3+k_3)$ and looked for possible integer values of $k_i$ that would satisfy it. What I posted then was the resulting solution. I couldn't find a general method of that type for solving the k's analytically such that it would work for any given number of a certain kind.

Anonymous wrote: I had a hunch that since it had to be proven that it was composite that its primes were somehow related to the the other numbers. We can strengthen the above into a "criteria" to generate problems similar to this. Anonymous' Criterion wrote: The resulting divisors to be found each have constant distance from one of the original three "large numbers" If the problem was designed to be solved without a calculator, the above would be useful to at least check for a solution. Take $ p_i$ as before. Generalize the constant term $ 320$ to $ k$. Define $ r_i$ as reasonable numbers such that\[ p_1p_2p_3+k=r_1r_2r_3\] Then, to find a linear relation we subtract the quantity $ p_1$ from each of the subscripted variables and suppose equality still holds. Then, $ k=(r_1-p_1)(r_2-p_1)(r_3-p_1)$ But $ p_1$ was arbitrarily chosen; thus, similar equations holds for $ p_2,p_3$. Specifically in this problem, we can find a quicker solution by looking at divisors of $ 320$: \[ 320=2*8*20=-10*-4*8=-16*-10*2\] Where for each $ i$, $ a_i,b_i,c_i$ are in arithmetic progression. We could formalize these observations (via interpolation) into a proof of the identity: \[ a(a+12)(a+18)+320=(a+2)(a+8)(a+20)\]simply by noting that we have found three values of $ a$ for which it holds, and the coefficients of $ a^3$ on either side are equal

If we denote $x=991$ we get: $(x-2)(x+10)(x+16)+320$ $=x^3+24x^2+108x$ $=x(x+6)(x+18)$ $=991\cdot 997\cdot 1009$ Since all three numbers are prime, we are done.