Christyan Tamaro's integer pairs

Without loss of generality,assume $a\geq b$ .Then $\frac{4b-1}{a}<\frac{4a}{a}=4$ .But $\frac{4b-1}{a}$ must be odd,since it divides an odd number $4b-1$ .Then there are 2 cases: Case 1: $\frac{4b-1}{a}=1$ In this case $b|4a-1=16b-5 \Rightarrow b|5$ .Thus,the solutions in this case are $(3,1)$ and $(19,5)$ .

Case 2: $\frac{4b-1}{a}=3$ In this case the ordered pairs that satisfy this is in form $(4k+1,3k+1)$ .Then $3k+1=b|4a-1=16k+3 \Rightarrow 3k+1|16(3k+1)-3(16k+3)=7$ .We have $k=0$ and $k=2$ ,yielding the solutions $(1,1)$ and $(9,7)$ .

Since in 3 of the solutions we can flip $a$ and $b$ ,then in total there are $1+2\cdot3=7$ solutions

Since $\frac {4a-1}{b}$ , $\frac {4b-1}{a}$ are integers, it follows that $\frac {4a-1}{b}\cdot\frac {4b-1}{a}$ would also be an integer. This then expands to give us $16-\frac{4}{a}-\frac{4}{b}+\frac{1}{ab}=16-(\frac{4}{a}+\frac{4}{b}-\frac{1}{ab})$ .

It thus suffices to find ordered pairs $(a,b)$ where $\frac{4}{a}+\frac{4}{b}-\frac{1}{ab}$ is an integer. When $a=1$ , we have to find $b$ where $\frac{3}{b}$ is an integer, which occurs when $b=1,3$ . When $a=3$ , we have to find $b$ where $\frac{11}{b}$ and $\frac{4b-1}{3}$ are both integers, which is an integer only when $b=1$ (basically an inverse of one of our previous cases)

So, if we have $a \geq 5$ , we must have $b \geq 5$ (If we have $b<5$ , then following the above cases, we would also have $a<5$ .). With $a,b \geq 5$ , we have $\frac{4}{a}+\frac{4}{b}-\frac{1}{ab}<\frac{4}{4}+\frac{4}{4}=1+1=2$ which means that we must have $\frac{4}{a}+\frac{4}{b}-\frac{1}{ab}=1$ .

Multiplying our expression above by $ab$ , we get $4a+4b-1=ab$ . Rearranging it gives us $ab-4a-4b+16=(a-4)(b-4)=16+(-1)=15$ The ordered pairs $(a,b)$ we can get from this are then $(5,19),(7,9),(9,7),(19,5)$ , basically $(a-4)$ and $(b-4)$ being positive factors of $15$ . With the 3 above solutions ( $(a,b)=(1,1),(1,3),(3,1)$ ), there are altogether $\fbox{7}$ such ordered pairs.

It is easy to see that both of $a$ and $b$ are odd integers only.

(For instant, if $a$ is even, then $2 \mid a$ and $a \mid (4b-1)$ imply $2 \mid (4b-1)$ . Consequently, $2 \mid -1$ , which is impossible.)

Since $a \mid (4b-1)$ and $b \mid (4a-1)$ , $ab \mid (4a-1)(4b-1)$ .

Which implies $ab \mid (4a+4b-1)$ .

Therefore $ab \le |4a+4b-1|$ .

Because $a,b \ge 1$ , $4a+4b-1 >0$ .

Hence $ab \le 4a+4b-1$ .

Factorize and obtain $(a-4)(b-4) \le 15$ .

This gives us the result that if $b>4$ then $a \le 19$ .

Thus we can consider case-by-case where $b=1,3$ and $a=1,3,5,7,9,11,13,15,17,19$ while $b>4$ .

Case I : $b=1,3$ .

If $b=1$ we get $a \mid 3$ and $1 \mid (4a-1)$ . Then $(a,b)=(1,1),(3,1)$ .

If $b=3$ we get $a \mid 11$ and $3 \mid (4a-1)$ . Then $(a,b)=(1,3)$ .

Case II : $a=1,3,5,7,9,11,13,15,17,19$ and $b>4$ .

If $a=1$ we get $b \mid 3$ and $1 \mid (4b-1)$ . This case is impossible.

If $a=3$ we get $b \mid 11$ and $3 \mid (4b-1)$ . This case is impossible.

If $a=5$ we get $b \mid 19$ and $5 \mid (4b-1)$ . Then $(a,b)=(5,19)$ .

If $a=7$ we get $b \mid 27$ and $7 \mid (4b-1)$ . Then $(a,b)=(7,9)$ .

If $a=9$ we get $b \mid 35$ and $9 \mid (4b-1)$ . Then $(a,b)=(9,7)$ .

If $a=11$ we get $b \mid 43$ and $11 \mid (4b-1)$ . This case is impossible.

If $a=13$ we get $b \mid 51$ and $13 \mid (4b-1)$ .This case is impossible.

If $a=15$ we get $b \mid 59$ and $15 \mid (4b-1)$ . This case is impossible.

If $a=17$ we get $b \mid 67$ and $17 \mid (4b-1)$ . This case is impossible.

If $a=19$ we get $b \mid 75$ and $19 \mid (4b-1)$ . Then $(a,b)=(19,5)$ .

From all cases, we conclude that $(a,b)=(1,1),(1,3),(3,1),(5,19),(7,9),(9,7),(19,5)$ .

Therefore, there are $7$ ordered pairs satisfying the problem.

If $a=b$ , it is easy to check $(1,1)$ is the only solution.

Let $a<b$ , then $(4a-1)$ which is odd can be either of $b$ or $3b$ . (not higher since $a<b$ )

Case 1: If $4a-1=b$ , then from the other relation $a| (16a-5) \Rightarrow a|5 \Rightarrow a=1,5$ . This contributes to solutions $(1,3),(5,19)$ .

Case 2: If $4a-1=3b$ , then from the other relation $3a| (16a-7) \Rightarrow 3a|(7-a)$ . Now we can have $7-a = 0,3a,6a,$ etc. It is easy to check integral solutions exist iff $(7-a)=0,6a \Rightarrow a=1,7$ This contributes to solution $(7,9)$ [ $(1,1)$ is already counted].

Since the pairs are ordered, for the latter three, we have 6 permissible pairs along with first one, giving 7 solutions in all.

First let us find solutions where a=b.

In this case we have a|4a-1 which implies a|-1.

Since a is a positive integers, a must be 1.

So only solution with a=b is (a,b)=(1,1).

Now we shall find solutions where a is not equal to b.

Assume that that a < b.

Since the given fact is symmetric in a and b, we don't need to consider the case b < a.

If (a,b)=(a 0, b 0) is a solution in the first case, (a,b)=(b 0,a 0) will be a solution in the later case and vice versa.

So we have a divisor of 4a-1 namely b which is greater than a.

Suppose 4a - 1 = bx. Since b > a, x < 4.

x can not be 2 since 4a-1 is odd. So x can only be 1 or 3.

Case I) x = 1, i.e., 4a - 1 = b.

In this case from a|4b-1, we get a|5. So a = 1 or a = 5.

If a = 1, we get b = 3.

If a = 5, we get b =19.

So we get the solutions (1,3) and (5,19) from this case.

Case II) x = 3, i.e., 4a - 1 = 3b.

In this case from a|4b-1, we get a|7. So a = 1 or a = 7.

If a = 1 we get b = 1 which is impossible since a<b.

If a = 7 we get b = 9.

So from this case we get the solution (7,9).

So all possible solutions are (1,1), (1,3), (5,19), (7,9), (3,1), (19,5), (9,7).

With the lazy hacker's method. This problem had me for 4 days, in the end I just decided to brute force it with a simple algorithm.

def areIntegerSolutions(a,b):
  return (((4*a)-1) % b == 0) and  (((4*b)-1) % a == 0)

for a in range(1,10000):
  for b in range(1,10000):
    if areIntegerSolutions(a,b):
      print "("+str(a)+","+str(b)+")\n"

>>> (1,1)

>>> (1,3)

>>> (3,1)

>>> (5,19)

>>> (7,9)

>>> (9,7)

>>> (19,5)