Problem 687

Since the left side of the given equation is a multiple of 6 , then the right side, $c^{2}$, is also a multiple of 6 .

Since $c^{2}$ is a multiple of 6 , then $c^{2}$ is a multiple of 2 and a multiple of 3 .

Since 2 and 3 are different prime numbers, then the positive integer $c$ itself must be a multiple of 2 and a multiple of 3 . This is because if $c$ is not a multiple of 3 , then $c^{2}$ cannot be a multiple of 3 , and if $c$ is not even, then $c^{2}$ cannot be even.

Therefore, $c$ is a multiple of each of 2 and 3 , and so is a multiple of 6 .

Thus, there are five possible values for $c$ in the given range: $6,12,18,24,30$.

If $c=6$, then $6 a b=36$ and so $a b=6$.

Since $1 \leq a<b<6$ (because $c=6$ ), then $a=2$ and $b=3$.

If $c=12$, then $6 a b=144$ and so $a b=24$.

Since $1 \leq a<b<12$, then $a=3$ and $b=8$ or $a=4$ and $b=6$.

(The divisor pairs of 24 are $24=1 \cdot 24=2 \cdot 12=3 \cdot 8=4 \cdot 6$. Only the pairs $24=3 \cdot 8=4 \cdot 6$ give solutions that obey the given restrictions, since in the other two pairs, the larger divisor does not satisfy the restriction of being less than 12 .)

If $c=18$, then $6 a b=324$ and so $a b=54$.

Since $1 \leq a<b<18$, then $a=6$ and $b=9$.

(The divisor pairs of 54 are $54=1 \cdot 54=2 \cdot 27=3 \cdot 18=6 \cdot 9$.)

If $c=24$, then $6 a b=576$ and so $a b=96$.

Since $1 \leq a<b<24$, then $a=6$ and $b=16$ or $a=8$ and $b=12$.

(The divisor pairs of 96 are $96=1 \cdot 96=2 \cdot 48=3 \cdot 32=4 \cdot 24=6 \cdot 16=8 \cdot 12$.)

If $c=30$, then $6 a b=900$ and so $a b=150$.

Since $1 \leq a<b<30$, then $a=6$ and $b=25$ or $a=10$ and $b=15$.

(The divisor pairs of 150 are $150=1 \cdot 150=2 \cdot 75=3 \cdot 50=5 \cdot 30=6 \cdot 25=10 \cdot 150$.)

Therefore, the triples $(a, b, c)$ of positive integers that are solutions to the equation $6 a b=c^{2}$ and that satisfy $a<b<c \leq 35$ are

$$(a, b, c)=(2,3,6),(3,8,12),(4,6,12),(6,9,18),(6,16,24),(8,12,24),(6,25,30),(10,15,30)$$

There are 8 such triplets.