a1: true, difference of two squares, (319-318)(319+318)
a2: false, sum of two cubes, 125+512
b1: false, not even
b2: true, not semiprime, 7*7*13
c1: true, sum of two squares, 196+441
c2: false, not both a square and a cube
d1: true, exactly 3 prime factors (if the 7 is counted twice)
d2: false, not prime
e1: false, not the sum of a square and a cube
e2: true, sum of 3 squares, 25+36+576
To see why e1 is false, I had to check that none of 637-a^3 are squares for a=1,2,3,4,5,6,7,8.
Reasoning: one of d1 and d2 is true, so b2 must be true. Therefore b1 is false, and the number is odd. Every odd number is the difference between two squares, 2k+1 = (k+1)^2 – k^2, so a1 is true, a2 is false, and the number is a sum of two cubes.
To get an odd sum of two cubes we have to add an odd cube to an even cube. The 3-digit numbers of this form are: 217, 513, 243, 539, 133, 189, 341, 637, 351, 407, 559, 855, 737, 793, 945.
None of these is a perfect 6th power, so c2 is false and we want a sum of two squares. A number is a sum of two squares exactly when every prime that’s 3 mod 4 occurs with even exponent in its prime factorization. This narrows it down to just 637 and 793. 793=13*61 is a semiprime, so 637 is the answer.
That line of reasoning doesn’t use e at all. Anyone have a different solution?
Not at all different. But the line of reasoning is nice. I have had this problem for over a decade, and never noticed that e. was superfluous. Thanks for pointing it out!
I tried solving it completely differently, mostly because I thought “three prime factors” meant “three distinct prime factors,” in which case the smallest number was 5*13*17 = 1105.
I’m getting 637:
a1: true, difference of two squares, (319-318)(319+318)
a2: false, sum of two cubes, 125+512
b1: false, not even
b2: true, not semiprime, 7*7*13
c1: true, sum of two squares, 196+441
c2: false, not both a square and a cube
d1: true, exactly 3 prime factors (if the 7 is counted twice)
d2: false, not prime
e1: false, not the sum of a square and a cube
e2: true, sum of 3 squares, 25+36+576
To see why e1 is false, I had to check that none of 637-a^3 are squares for a=1,2,3,4,5,6,7,8.
Reasoning: one of d1 and d2 is true, so b2 must be true. Therefore b1 is false, and the number is odd. Every odd number is the difference between two squares, 2k+1 = (k+1)^2 – k^2, so a1 is true, a2 is false, and the number is a sum of two cubes.
To get an odd sum of two cubes we have to add an odd cube to an even cube. The 3-digit numbers of this form are: 217, 513, 243, 539, 133, 189, 341, 637, 351, 407, 559, 855, 737, 793, 945.
None of these is a perfect 6th power, so c2 is false and we want a sum of two squares. A number is a sum of two squares exactly when every prime that’s 3 mod 4 occurs with even exponent in its prime factorization. This narrows it down to just 637 and 793. 793=13*61 is a semiprime, so 637 is the answer.
That line of reasoning doesn’t use e at all. Anyone have a different solution?
Not at all different. But the line of reasoning is nice. I have had this problem for over a decade, and never noticed that e. was superfluous. Thanks for pointing it out!
I tried solving it completely differently, mostly because I thought “three prime factors” meant “three distinct prime factors,” in which case the smallest number was 5*13*17 = 1105.