Prove a is a composite number

[anemone]

Positive integers $a$ and $b$ are such that $\dfrac{5a^4+a^2}{b^4+3b^2+4}$ is an integer. Prove that $a$ is composite.

 

[jvicens]

Thank you for bringing up this interesting problem. I am always eager to explore and solve mathematical challenges.

To prove that $a$ is composite, we need to show that it is not a prime number. Let us assume that $a$ is a prime number. This means that it only has two factors, 1 and itself. Therefore, $a^2$ and $a^4$ will also be prime numbers, as they are just the product of two factors of $a$.

Now, let us look at the expression $\dfrac{5a^4+a^2}{b^4+3b^2+4}$. We can rewrite it as $\dfrac{a^2(5a^2+1)}{b^2(b^2+3)+4}$. Since $a^2$ is a prime number, it cannot be divided by any other prime number except for 1 and itself. Therefore, in order for the entire expression to be an integer, the numerator $(5a^2+1)$ must also be divisible by $a^2$. This means that $5a^2+1 = ka^2$, where $k$ is some integer.

Rearranging this equation, we get $a^2(k-5) = 1$. Since $a^2$ and $k-5$ are both integers, this can only happen if $k-5 = 1$, which means that $k = 6$. Substituting this back into our original equation, we get $5a^2+1 = 6a^2$, which simplifies to $a^2 = 1$. But, we know that $a$ is a positive integer, so $a^2$ cannot be equal to 1. This means that our assumption that $a$ is a prime number is incorrect.

Therefore, we have proven that $a$ cannot be a prime number and must be a composite number. This also means that $a^2$ and $a^4$ are not prime numbers, and can be divided by other prime numbers. Thus, the entire expression $\dfrac{5a^4+a^2}{b^4+3b^2+4}$ will also be divisible by these prime numbers, making it an integer.

I hope this explanation helps to clarify and solve the problem.