MHB Solve for positive integer solutions

[anemone]

Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

 

[Greg]

Find the positive integer numbers $a$ and $b$ such that $65(a^3b^3 + a^2 + b^2) = 81(ab^3 + 1)$.

Spoiler

$$65(a^3b^3+a^2+b^2)=81(ab^3+1)$$

$$65a^2(ab^3+1)+65b^2=81(ab^3+1)$$

$$(81-65a^2)(ab^3+1)=65b^2\implies a=1\text{ if there are solutions in the positive integers.}$$

$$16b^3-65b^2+16=0\implies b=4\text{ if }b\text{ is an integer.}$$

$$(a,b)=(1,4)\text{ are the only positive integers that solve the given equation.}$$

 

[anemone]

Spoiler

$$65(a^3b^3+a^2+b^2)=81(ab^3+1)$$

$$65a^2(ab^3+1)+65b^2=81(ab^3+1)$$

$$(81-65a^2)(ab^3+1)=65b^2\implies a=1\text{ if there are solutions in the positive integers.}$$

$$16b^3-65b^2+16=0\implies b=4\text{ if }b\text{ is an integer.}$$

$$(a,b)=(1,4)\text{ are the only positive integers that solve the given equation.}$$

Very good job, greg1313!

My solution:

Spoiler

The given equation can be rewritten as $b^3(a)(65a^2-81)+65b^2+(65a^2-81)=0$ (*).

We have two cases to consider here, first is $65a^2-81>0$, which leads to $a≥2$. Replacing these two important information into (*) we see that we have the relation where it says always positive+always positive+always positive=0, which tells us there is no solution for $a≥2$.

That also means we only have to consider for $0≤a≤1$. Now, solving the above equation for $b$ when $a=1$, we find $b=4$ and there is no solution for the case $a=0$ and we are hence done.

 

[kaliprasad]

Very good job, greg1313!

My solution:

Spoiler

Now, solving the above equation for $b$ when $a=1$, we find $b=4$ and there is no solution for the case $a=0$ and we are hence done.

Spoiler

because a is positive we need not consider $a = 0$