Review question

# Can we find integers that satisfy $a^3 + 3b^3 = 9c^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7096

## Solution

1. Suppose that $a$, $b$ and $c$ are integers that satisfy the equation $a^3 + 3b^3 = 9c^3.$

Explain why $a$ must be divisible by $3$, and show further that both $b$ and $c$ must also be divisible by $3$. Hence show that the only integer solution is $a = b = c = 0$.

We have $a^3 = 9c^3 - 3b^3 = 3(3c^3 - b^3),$ so $a^3$ is a multiple of $3$.

Now if $3$ does not divide $a$, it does not divide $a^3$, so $3$ must divide $a$. Thus we can write $a = 3d$, where $d$ is an integer, and we have $(3d)^3 + 3b^3 = 9c^3,$ which, on dividing by $3$, gives $9d^3 + b^3 = 3c^3.$

By the same argument, as $b^3 = 3(c^3 - 3d^3),$ it follows that 3 divides $b^3$, and also $b$. We repeat the same trick, writing $b = 3e$, where $e$ is an integer, so that $9d^3 + (3e)^3 = 3c^3.$

We again divide by $3$ to get $3d^3 + 9e^3 = c^3,$ so that 3 divides $c^3$, and also $c$. We then write $c = 3f$, where $f$ is an integer, giving $3d^3 + 9e^3 = (3f)^3.$

Finally, we divide this equation by $3$ to get $d^3 + 3e^3 = 9f^3.$

Now this is the same equation that we started with. So if $a$, $b$, $c$ are integers which satisfy the equation, then so are $d = \dfrac{a}{3}$, $e = \dfrac{b}{3}$ and $f = \dfrac{c}{3}$.

Thus from our imagined starting solution $(a,b,c)$, if $(a,b,c)$ is not $(0,0,0)$, we have constructed a smaller integer solution $(d,e,f)$.

We could repeat this process indefinitely, but there are only finitely many integers smaller than $a$; a contradiction!

The only way out is if $(a,b,c) = (0,0,0)$, when $(d,e,f)$ is not a smaller solution, but is $(0,0,0)$ also.

More precisely once again…

if $a$, $b$, $c$ are integers which satisfy the equation, then so are $d = \dfrac{a}{3}$, $e = \dfrac{b}{3}$ and $f = \dfrac{c}{3}$, so that for any $n$, the numbers $\frac{a}{3^n}$, $\frac{b}{3^n}$ and $\frac{c}{3^n}$ are also integers which satisfy the equation.

But if $\frac{a}{3^n}$ is an integer for all $n \ge 0$, we must have $a = 0$, and similarly for $b$ and $c$.

Therefore the only integer solution is $a = b = c = 0$.

1. Suppose that $p$, $q$ and $r$ are integers that satisfy the equation $p^4 + 2q^4 = 5r^4.$

By considering the possible final digit of each term, or otherwise, show that $p$ and $q$ are divisible by $5$. Hence show that the only integer solution is $p = q = r = 0$.

We are interested in the equation $p^4 + 2q^4 = 5r^4.$

This table gives $p^4 + 2q^4 \mod 5$, that is, the remainder when $p^4 + 2q^4$ is divided by 5.

The only way $p^4 + 2q^4$ can be $0$ mod $5$ is if $p = 0 \mod 5$, and $q = 0 \mod 5$, in other words, if $5$ divides both $p$ and $q$.

Now as in part (i), we write $p = 5a$ and $q = 5b$, where $a$ and $b$ are both integers, to get $(5a)^4 + 2(5b)^4 = 5r^4.$

Dividing both sides by $5$ gives $5^3a^4 + 2(5^3b^4) = r^4,$

so as before, $r^4$ must be a multiple of $5$, and since $5$ is prime, $r$ must be a multiple of $5$. Then writing $r = 5c$ gives $5^3a^4 + 2(5^3b^4) = (5c)^4,$ which yields $a^4 + 2b^4 = 5c^4$ on dividing by $5^3$.

So once again, if $p$, $q$, $r$ give an integer solution to the equation, then so do $a = \dfrac{p}{5}$, $b = \dfrac{q}{5}$ and $c = \dfrac{r}{5}$. Repeating this, so are $\dfrac{p}{5^n}$, $\dfrac{q}{5^n}$, $\dfrac{r}{5^n}$ for any integer $n \geq 0$, and as before, this shows that the only integer solution is $p = q = r = 0$.

The type of argument used here (using one imagined solution to find a smaller one) is known as infinite descent, which was popularised by Pierre de Fermat.