jmc

Question

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form

P (x) = 2 x^{3} - 2 a x^{2} + (a^{2} - 81) x - c

for some positive integers

a

and

c

. Can you tell me the values of

a

and

c

?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of

a

." He writes down a positive integer and asks, "Can you tell me the value of

c

?"

Jon says, "There are still two possible values of

c

."

Find the sum of the two possible values of

c

.

Accepted Answer

Let r, s, t be the three positive integer roots of P(x). Then by Vieta's formulas, aligned r+s+t &= a, rs+st+rt &= a^2-812, rst &= c2. alignedSubstituting the first equation into the second to eliminate a, we have rs+st+rt = (r+s+t)^2 - 812 = (r^2+s^2+t^2) + 2(rs+st+rt) - 812.This simplifies to r^2 + s^2 + t^2 = 81.Therefore, each of r, s, t lies in the set 1, 2, , 9. Assuming without loss of generality that r s t, we have 81=r^2+s^2+t^2 3t^2, so t^2 27, and t 6. We take cases: If t = 6, then r^2+s^2 = 81 - 6^2 = 45; the only solution where r s 6 is (r, s) = (3, 6). If t = 7, then r^2+s^2 = 81-7^2 = 32; the only solution where r s 7 is (r, s) = (4, 4). If t = 8, then r^2+s^2 = 81-8^2 = 17; the only solution where r s 8 is (r, s) = (1, 4). Therefore, the possible sets of roots of such a po…