jmc

Question

For how many positive integers p does there exist a triangle with sides of length 3p-1, 3p, and p^2 + 1?

Accepted Answer

Applying the Triangle Inequality, we have that

(3 p - 1) + (3 p) > p^{2} + 1,

so

0 > p^{2} - 6 p + 2.

We can find the values of

p

that satisfy this inequality by completing the square. Adding 7 to both sides gives

7 > p^{2} - 6 p + 9

, so

7 > (p - 3)^{2}

. Since

p

must be a positive integer, the only possible values of

(p - 3)^{2}

are 0, 1, and 4. Therefore, the possible values of

p

are 1, 2, 3, 4, and 5. Let's find

(3 p - 1, 3 p, p^{2} + 1)

for each possible

p

:

If

p = 1,

then

(3 p - 1, 3 p, p^{2} + 1) = (2, 3, 2) .

If

p = 2,

then

(3 p - 1, 3 p, p^{2} + 1) = (5, 6, 5) .

If

p = 3,

then

(3 p - 1, 3 p, p^{2} + 1) = (8, 9, 10) .

If

p = 4,

then

(3 p - 1, 3 p, p^{2} + 1) = (11, 12, 17) .

If

p = 5,

then

(3 p - 1, 3 p, p^{2} + 1) = (14, 15, 26) .

All of these look good, so we see that there are

5

possibilities for

p .