Problem:
Are there any triples (a,b,c) of positive integers such that (a−2)(b−2)(c−2)+12 is a prime that properly divides the positive number a2+b2+c2+abc−2017 ?
Solution:
Suppose (a,b,c) is such a triple. The prime (a−2)(b−2)(c−2)+12 also divides
​a2+b2+c2+abc−2017−(a−2)(b−2)(c−2)−12=(a+b+c)2−4(a+b+c)+4−2025=(a+b+c−2)2−452=(a+b+c−47)(a+b+c+43)​
We may assume without loss of generality that a≤b≤c. If a=b=1,c+10 must be a prime that properly divides c2+c−2015, implying c+10 divides 1925=52⋅7⋅11. So c+10=11, and we obtain the triple (1,1,1). However, this does not make a2+b2+c2+abc−2017 positive.
If a=1 and b=2, then (a−2)(b−2)(c−2)+12=12 is not prime. If a=1 and b=3,14−c must be a prime. The allowable choices for c are 3,7,9,11 and 12 , but none of these work. If a=1 and b=4, the prime is even, so must be 2 and hence c=7, but this doesn't work either. If a=1 and b≥5 then c≥5 also, so (a−2)(b−2)(c−2)+12≤12−9=3, and the only possibility is b=c=5, but this also doesn't work. This rules out the cases with a=1. Also a=2 is impossible, again because 12 is not prime.
Now let x=a−2,y=b−2,z=c−2. We now know that 1≤x≤y≤z and (x+2)+(y+ 2) +(z+2)>47. So x+y+z≥41, and therefore z≥14. The prime xyz+12 cannot divide (x+2)+(y+2)+(z+2)−47 since xyz−4>x+y+z−41. Indeed, this latter inequality reduces to x(yz−1)>y+z−37, which will follow if we can prove that yz−1>y+z−37 (since x≥1 ). The last statement is equivalent to (y−1)(z−1)>−36, which is evidently true.
Hence xyz+12 divides (x+2)+(y+2)+(z+2)+43. They cannot be equal: x,y,z must all be odd, otherwise xyz+12 is not prime, but then (x+2)+(y+2)+(z+2)+43 is even and so not equal to xyz+12. Thus 2(xyz+12)≤x+y+z+49, implying 2yz−1≤x(2yz−1)≤y+z+25. It follows that (2y−1)(2z−1)≤53. Earlier we proved that z≥14; since z is odd, we must in fact have z≥15. Moreover, 2y−1≤53/(2z−1)≤53/29<2. Therefore x=y=1. It follows that z+12 is prime and 15≤z≤27; therefore z=17, 19, or 25 . Also, z+12 divides (x+2)+(y+2)+(z+2)+43=z+51. However, this is false for z=17,19, or 25 . Consequently, the answer is negative; i.e., the requested triples (a,b,c) do not exist.
The problems on this page are the property of the MAA's American Mathematics Competitions