To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer

Question

How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and

$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ?

Answer 1

If $(a,b)$ is an ordered pair of positive integers such that $\gcd(a,b)=1$ and $$\frac{a}{b}+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab},$$ is an integer, then $9ab$ divides $9a^2+14b^2$. In particular $a$ and $b$ both divide $9a^2+14b^2$, and so $a$ divides $14b^2$ and $b$ divides $9a^2$. Because $\gcd(a,b)=1$ it follows that $a$ divides $14$ and $b$ divides $9$. If $b=9$ then $81a$ divides $9a^2+14\cdot 81$, so $81$ divides $9a^2$ and hence $3$ divides $a$, contradicting the fact that $\gcd(a,b)=1$. Of course $b\neq1$ because $9$ does not divide 14. Hence $b=3$, and we conclude that there are precisely four such pairs.

Answer 2

We are looking for $9ab\ |\ 9a^2+14b^2$.

Now, as everything else is divisible by $9$, we immediately get that $9\ |\ b^2$, i.e., $3\ |\,b$.

Similarly, looking at divisibility by $a$, we get $a\,|\,14$. This gives only four possibility on $a$: it can be $\ 1,\ 2,\ 7,\ 14$.

If we write $b=3k$, we get $27ak\ |\ 9a^2+14\cdot 9k^2$, that simplifies to $$3ak\ |\ a^2+14k^2\,.$$

Now, by divisibility by $k$, we have $k\,|\,a^2$, which is only possible if $k=1$ by condition $\gcd(a,b)=1$.

From this there are only $4$ possibilities, check them manually.

Answer 3

$$\frac ab+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab}$$

If this expression is an integer, since $a$ and $b$ are coprime, $a$ divides $14$ and $b$ divides $9$.

If $b=1$, we get $$a+\frac{14}{9a}$$ which is not an integer.

If $b=9$, $$\frac{a}9+\frac{14}a=\frac{a^2+126}{9a}$$ which is neither an integer, since $a$ can't be a multiple of $9$.

If $b=3$, $$\frac a3+\frac{14}{3a}=\frac{a^2+14}{3a}$$ which is an integer for every possible value of $a$ (namely, $1$, $2$, $7$ and $14$).

So there is $4$ ordered pairs of positive integers.

Answer 4

I first find out what the values of a/b are which make the expression an integer, then check which of the solutions are rational numbers. Let t = a/b, then t + 14/(9t) should be an integer.

Solve t + 14/(9t) = n: t^2 + 14/9 = nt, t^2 - nt + n^2/4 = n^2/4 - 14/9. t - n/2 = +/- sqrt (n^2/4 - 14/9).

t should be rational, so n^2/4 - 14/9 should be the square of a rational number. Multiply by 36, and 9n^2 - 56 should be the square of a rational number, which means it must be the square of an integer.

When n >= 10, (3n-1)^2 < 9n^2 - 56 < (3n)^2, so 9n^2 - 56 cannot be the square of an integer. n = 1 to n = 9 can be checked manually, and we find that n = 3 makes 9n^2 - 56 = 25 = 5^2, n = 3 makes 9n^2 - 56 = 169 = 13^2.

n = 3: t - 3/2 = +/- sqrt (9/4 - 14/9) = +/- sqrt (25/36) = +/- 5/6; t = 14/6 or t = 4/6.

n = 5: t - 5/2 = +/- sqrt (25/4 - 14/9) = +/- sqrt (169/36) = +/- 13/6, t = 28/6 or t = 2/6.

Together: a/b is one of 7/3, 2/3, 14/3 or 1/3

Checking: 7/3 + (14*3 / 9*7) = 7/3 + 2/3 = 3 2/3 + (14*3 / 9*2) = 2/3 + 7/3 = 3 14/3 + (14*3 / 9*14) = 14/3 + 1/3 = 5 1/3 + (14*3 / 9*1) = 1/3 + 14/3 = 5.

Answer 5

A slight generalization shows more clearly the structure behind the number of solutions:

Theorem $\ $ Suppose that $\,c\,$ is a positive integer and $\,p\,$ is prime such that $\,p^2\nmid c.\ $ Then

$\ \ $ there are coprime integers $\, a,b > 0\,$ with $\,\dfrac{a}{b} +\dfrac{b\,c\ }{a\,p^2}\in\Bbb Z\iff b=p,\,\ c = aa',\,\ p\mid a\!+\!a' $

Therefore the #solutions = #factorizations $\,c\,$ into two factors $\,>0\,$ with sum divisible by $\,p.$

Proof $\ $ By Euclid $\,(a,b)=1\,\Rightarrow\, (a,b^2)=1=(b,a^2),\ $ thus $\ abp^2\mid p^2a^2 + cb^2\,\Rightarrow\,a\mid c,\,\ b\mid p^2.\,$ Let $\ a' = c/a.\,$ By unique factorization $\,b\mid p^2\Rightarrow\,b = 1\,$ or $\,b=p\,$ or $\,b = p^2,\, $ yielding $3$ cases:

$\qquad\qquad\qquad b\, =\, 1\,\Rightarrow\,\ a + \dfrac{a'}{p^2}\in\Bbb Z\,\Rightarrow\, p^2\mid a'\mid c,\,$ contra hypothesis.

$\qquad\qquad\qquad b = p^2\,\Rightarrow\, \dfrac{a}{p^2}+ a'\in\Bbb Z\,\Rightarrow\, p^2\mid a\mid c,\,$ contra hypothesis.

$\qquad\qquad\qquad b\, =\, p\, \Rightarrow\, \dfrac{a}p + \dfrac{a'}{p}\in\Bbb Z\iff p\mid a+a'\quad $ QED

Yours is special case $\,p = 3,\,\ c = 14\,$ with factors $\,a,a' = 1,14;\,\ 2,7;\,\ 7,2;\,\ 14,1.$