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Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5=b^4$, $c^3=d^2$, and $c-a=19$. Determine $d-b$.

I know this question isn't particularly hard, but I've been having trouble getting any substantial ground. I've tried substituting that $c=a+19$ and $a=c-19$ into the first two equations, but it hasn't looked very useful.

A path to a solution or a few hints would be great. Thanks!

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We can say that $b$ divides $a$, and $d$ divides $c$. – Yiyuan Lee Jun 28 '14 at 15:13
up vote 3 down vote accepted

From $a^5 = b^4 \implies a = \left(\frac{b}{a}\right)^4$, we can deduce that $\frac{b}{a}$ is an integer. I will let $m = \frac{b}{a}$.

Similarly, from $c^3 = d^2 \implies c = \left(\frac{d}{c}\right)^2$, we can deduce that $\frac{d}{c}$ is an integer. I will let $n = \frac{d}{c}$.

Since $a = m^4$ and $c = n^2$, we have

$$n^2 - m^4 = 19$$ $$(n + m^2)(n - m^2) = 19$$

Since $19$ is a prime, we only have to check the cases for which $19 = 1\cdot19$ and for which $19 = -1\cdot-19$. If we let $n + m^2 = 19$ and $n - m^2 = 1$, then we can get $n = 10, m = 3$. I suppose it is trivial to show that all other cases do not give valid results.

Hence, $\frac{d}{c} = n = 10 \implies d = 10c$, and $\frac{b}{a} = m = 3 \implies b = 3a$. It should be easy to continue on from here!

(Just in case you wanna check: $a = 81, b = 243, c = 100, d = 1000$)

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Thank you! This solution was very clear and I understand the problem so much more now. :) – math-sd Jun 28 '14 at 15:42

Here are a few things worth noting. Observe that if $p$ is a prime number that divides $a$--say $a=p^km$ for some positive integers $k,m$ such that $p\not\mid m$--then $$b^4=a^5=p^{5k}m^5.$$ Hence, in particular, $4\mid5k,$ and so $4\mid k.$ (Why?) This is true for every such prime $p,$ and so $a$ is a perfect $4$th power of some integer $j$-- that is $a=j^4.$

By similar reasoning, $c=k^2$ for some integer $k$. It follows then that $$19=c-a=k^2-j^4=(k+j^2)(k-j^2).$$ Can you take it from here?

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Yes! Thank you. :) – math-sd Jun 28 '14 at 15:59

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