For how many pairs of distinct positive integers $a$ and $b$, both less than $100$, is $\dfrac{a}{b}$ the square of an integer?

I can only think of doing this question by casework on $a,b$, but I think that would get very tedious. Is there a better way?


Here is one approach.

  1. Note $a/b < 100$. Write out all perfect squares under 100.
  2. What are the ways of generating those perfect squares? Most are easy, like $64 = 64/1$ and $25 = 25/1 = 50/2 = 75/3$...

I can't think of a way to do it without having to use a bit of elbow grease (so to speak). This was my approach:

If $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ is an integer then $\sqrt{a} = c\sqrt{b}$ for some integer $c$. Thus $a = c^2b$ for some integer $c$. To find number of distinct integer pairs $a,b$ with $0<a<b<100$ and $a = c^2b$ we collect the perfect squares lying strictly between 1 and 100,


and for each square $c^2$ we count the multiples $c^2b$ such that $c^2b < 100$, then sum the counts.

Thus we have $$\lfloor\frac{99}{4}\rfloor+\lfloor\frac{99}{9}\rfloor+\lfloor\frac{99}{16}\rfloor+\lfloor\frac{99}{25}\rfloor+\lfloor\frac{99}{36}\rfloor+\lfloor\frac{99}{49}\rfloor+\lfloor\frac{99}{64}\rfloor+\lfloor\frac{99}{81}\rfloor \\ = 24+11+6+3+2+2+1+1 \\ = 50.$$


First approach:

We have $b=ak^2$, so if $a$ is fixed we just have to choose $k$ so that $k^2<\frac{100}{a}$

If $a$ is $1$ there are $9$ options for $k$.

If $a$ is $2$ there are $7$ options for $k$.

If $a$ is $3$ there are $5$ options for $k$..

If $a$ is $4,5$ or $6$ there are $4$ options for $k$ ($3$ numbers).

If $a$ is $7,8,9,10,11$ there are $3$ options for $k$ ($5$ numbers).

If $a$ is $12,...,$ or $24$ there are $2$ options for $k$ ($13$ numbers).

If $a$ is $25,\dots,$ or $99$ there is $1$ option for $k$ ($75$ numbers).

Therefore there are $9+7+5+4\cdot3+3\cdot5+2\cdot13+75=149$ pairs.

Second approach:

Suppose $b$ is fixed, then the number of options for $a$ is equal to the number of divisors of $b$ that are squares.

The number of such divisors is $1$ for all square free integers.

Notice a number less than $100$ cannot have two or more prime factors with multiplicity larger than $2$ (since the smallest such number is $2^25^2=100$.

So the number of square divisors for any number less than $100$ is:

  • $1$ if it is squarefree
  • $3$ if the number is $2^4,2^5,2^43,2^45,2^53,3^4$ ($6$ numbers)
  • $4$ if the number is $2^6,2^23^2,2^33^2$
  • $2$ in any other case.

Therefore the number of pairs is $99$ plus the number of non-squarefree numbers, plus $6$ plus $2\cdot3$.

How many non-squarefree numbers are there? First notice the $24$ that are multiples of $4$. There are also $11$ that are multiples of $9$. Of course, two of these where already counted, so $33$ up to here. Now add the ones that are multiples of $25$, this gives us another three, so we're up to $36$, finally, we must add $49$ and $98$, so in total there are $38$.

Therefore the final answer is : $99+38+6+6=149$

Edit: apparently you wanted $a\neq b$, this is easily fixable, just subtract $99$ from the answer, so it is $50$.


You have to approach this using casework. But there is a recognizable pattern. If $\frac{a}{b} = 4,$ then we have $\frac{100}{4} - 1 = 24.$ For $\frac{a}{b} = 9,$ we have $11$ ways. The summation is as follows: $$\sum_{n = 2}^{9}\lfloor\frac{99}{n^{2}}\rfloor.$$

  • $\begingroup$ I think you are forgetting to take a square root in your sum $\endgroup$ – Jorge Fernández Hidalgo Jan 19 '16 at 19:03
  • $\begingroup$ Why sqrt? There's no need, because we just count multiples @dREaM $\endgroup$ – K. Jiang Jan 19 '16 at 19:06
  • $\begingroup$ "Pairs of distinct integers", so a/b = 1 is not possible at all. $\endgroup$ – gnasher729 Jan 19 '16 at 19:36

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