Find $a$ for which $(a-3)(a-7)$ is a perfect square

Question

The only solutions seem to be 3 and 7 but I can't prove that there are no others.

Context: Find every value for integer a, for which $x^2-(a+5)x+5a+1$ expression can be factored as $(x+b)(x+c)$ where b and c are integers.

$x=\frac{a+5 \pm \sqrt{(a-3)(a-7)}}{2}$

(In a pre-calc book, btw. Real solutions only (don't know if it matters))

I have no idea where to start, forgive me for not having any real attempts.

Answer 1

To answer your question: when can $(a-3)(a-7)$ be a perfect square?

Suppose $$(a-3)(a-7)=k^2, $$ for some integer $k$. Then $$a^2-10a+21-k^2=0.$$ This implies that $$a = \dfrac{10\pm\sqrt{100-4(21-k^2)}}{2},$$ implying that $$a = \dfrac{10\pm 2\sqrt{4+k^2}}{2}.$$ As $a$ has to be an integer, we look for values of $k$ for which$\sqrt{4+k^2}$ is a perfect square. This is quite easy as $$k^2+4 = m^2 \implies 4 = (m-k)(m+k).$$ Following are the possibilities now:

$(m-k, m+k)=(\pm1,\pm4); (m-k, m+k)=(\pm2,\pm2); (m-k,m+k)=(\pm4,\pm1)$.

The first and last cases can be easily discarded as the solutions to which are not integers. Thus, we are left with the option $(m-k, m+k) = (\pm2,\pm2)$. Solving this, we would get $m=\pm2$ and $k=0$. Thus $$(a-3)(a-7)=0,$$ implying that $a=3, 7$ are the only solutions.

$\textbf{Note.}$ I really did not notice the comment of Will Jagy in the comments which runs on the similar lines as that of this. I just noticed it after writing this answer.

Answer 2

Alternative approach:

Let $~b = (a-5) \implies ~(a-3) \times (a-7) = b^2 - 4.~$

So, you are looking for some integer $~b,~$ other than $~b = \pm 2,~$ such that $~b^2 - 4~ = n^2 \implies b^2 - n^2 = 4,~$ for some integer $~n.$

If such a solution $~(b,n)~$ exists, then it must also exist with $~b,n \in \Bbb{Z_{\geq 0}},~$ with $~b > n.$

So, there must exist a $~k \in \Bbb{Z^+}~$ such that $~b = n+k \implies 4 = (n+k)^2 - n^2 = 2nk + k^2.~$

Since $~4~$ is an even number, $~k^2~$ must also be even, which implies that $~k~$ is even. So, the smallest possible value for $~k~$ is $~k = 2.~$

Then, this implies that $~4 \geq 2n(2) + (2)^2 = 4n + 4.~$

This requires that $~n = 0,~$ which implies that $~b^2 = 4 \implies b = \pm 2.~$ So, no solution other than $~b = \pm 2~$ is possible.

Answer 3

Find all integers $a$ such that $$x^2-(a+5)x+5a+1=(x+b)(x+c)$$ where $b,c$ are integers

The roots of the quadratic are $-b$ and $-c$, therefore $$(b+5)(c+5)=b c + 5(b+c)+25=5a+1+5(-a-5)+25=1$$

Either

$b+5=c+5=1$ in which case $b=c=-4$ and $a=3$

or

$b+5=c+5=-1$ in which case $b=c=-6$ and $a=7$