Prime number, square number. [closed]

Question

Let $a$ and $b$ be prime numbers such that $b\mid (a-1)$ and $a\mid(b^3-1)$. Prove that $a + b$ is a perfect square.

e.g. $(a,b)=(13,3)$ gives us $a+b=16=4^2$.

Answer 1

$b \mid a-1$

$b\leq a-1$

$a\mid b^3-1$

$a|(b-1)(b^2+b+1)$, but $a\geq(b+1)$ so $a|b^2+b+1$

Also $a=bk+1$

$bk+1\mid b^2+b+1$

$bk\leq b(b+1)$ and $b>0$ so $k\leq b+1$

$bk+1\mid b^3-1$

$bk+1\leq b^3-1$

$b(b^2-k)\geq 2$ so $b<k$

and finally $k=b+1$

$(a+b)=(b+1)^2$

Answer 2

Since $b \mid (a-1)$, we have $a=bk+1$ for some $k \in \mathbb{N}$. Now, since $a \mid b^3-1$, we have: $$(bk+1) \mid (b^3-1) \implies (bk+1) \mid (b^3-1)+(b^3k^3+1) \implies(bk+1) \mid b^3(k^3+1)$$ Clearly, since $bk+1$ is prime and $bk+1>b$, we have: $$(bk+1) \mid k^3+1 \implies k^3+1 \geqslant bk+1 \implies k \geqslant \sqrt{b}$$

We can easily see that $\sqrt{b} \not \in \mathbb{N}$ since $b$ is prime.

Now, let $k>\sqrt{b}$. Then, we have $k^2-b>0$.

$$(bk+1) \mid b^3-1 \implies (bk+1) \mid b^3+bk \implies (bk+1) \mid (b^2+k)$$ $$ \implies (bk+1) \mid (b^2k+k^2) \implies (bk+1) \mid (b^2k+b)+(k^2-b)$$ $$ \implies (bk+1) \mid b(b+k)+(k^2-b) \implies (bk+1) \mid (k^2-b)$$

Since $k^2-b>0$, we have $k^2-b \geqslant bk+1 \implies k>b$ since any value $k \leqslant b$ doesn't satisfy the inequality $k^2-b \geqslant bk+1$. We already know that $(bk+1) \mid (b^2+k)$ gives $b^2+k \geqslant bk+1$. Thus: $$b^2+k \geqslant bk+1$$ We have equality at $k=b+1$. Clearly, $k$ cannot be more, since the increment to the left side will be lesser than that to the right. This gives $k \leqslant b+1$. Remember that we also have $k>b$. This forces $k=b+1$.

Substituting $k=b+1$, we get $a=bk+1=b^2+b+1$. This works by identity. Thus:

$$a=b^2+b+1 \implies a+b=b^2+2b+1=(b+1)^2$$