I've worked out the reasoning, but how do I write the proof?

Question

This started our with a pretty trivial problem that went:

Fill in the blanks with whole numbers to make mathematically true statements. Do not use the same number twice within a statement.$$\frac{*}4+\frac{1}*=\frac{*}{20}$$

Now solutions were pretty easy, so I decided to change the problem and asked myself what solutions could be made when I must use a number twice. Solutions were easy for this format $$\frac{a}4+\frac{1}a=\frac{b}{20}$$ where $a\ne b$. But are there any solutions for the following format? $$\frac{b}4+\frac{1}a=\frac{a}{20}$$

To determine if there were any, I firstly rearranged the equation into a quadratic form, i.e.$$0=a^2-5ba+20$$ which yields solutions if $$a=\frac{5b\pm\sqrt{25b^2-80}}2$$Now this can only satisfy the condition of "whole numbers" if $\sqrt{25b^2-80}$ is a whole number. (Even then there is more that needs to be satisfied, so this is a minimal condition.) At this point, I didn't know how to prove this formally, so I decided to use excel to determine ${25b^2-80}$ for different values of b, and then use the vlookup function to find the nearest square, $n^2$, below that value. I then subtracted these two values because I figured that I was looking for any instances where $$\delta=25b^2-80-n^2\equiv0$$ Now I didn't find any, however, I found an unexpected pattern for the difference, $\delta$ given b=2, 3, ...

The value of $$\delta = (4, 1, 31, 16, 36, 56, 76, 9, 19, 29, 39, ...)$$ That is, for $b>8$, $$\delta =10(b-8)+9$$

I therefore have two questions. Why did this pattern emerge for $\delta$? And how do you formally write this reasoning, which does show that no value of "$a$" exists that is a whole number?

Answer 1

You have several choices already for the proof, so I'll focus on the pattern of values of $\delta.$ I think this is easier to reason about if we write the expression under the radical as $(5b)^2 - 80.$ In order for the radical to be a whole number, then, we would need the difference between two squares to be $80,$ where one of the squares is a square of a multiple of $5.$

For small values of $b$ there can be one or more squares strictly between $(5b)^2 - 80$ and $(5b)^2,$ so $\delta$ ends up being the difference between $(5b)^2$ and $(5b - n)^2$ where $n \geq 2.$ But

$$ (5b)^2 - (5b - 1)^2 = (5b)^2 - ((5b)^2 - 2(5b) + 1) = 10b - 1, $$

so if $b \geq 9$ then the difference between $(5b)^2$ and the next smaller square is at least $89,$ which is greater than $80.$ Hence there are no squares at all in the numbers from $(5b)^2 - 80$ to $(5b)^2$ (other than $(5b)^2$ itself), so $\delta$ is just the difference between $(5b)^2 - 80$ and the next smaller square, which is $(5b - 1)^2$:

\begin{align} \delta &= ((5b)^2 - 80) - (5b - 1)^2 \\ &= (10b - 1) - 80 \\ &= 10(b - 8) - 1\\ &= 10(b - 9) + 9. \end{align}

(Note that this is slightly different from the formula written in the question.)

By the way, since we found while doing this that there are no squares at all in the numbers from $(5b)^2 - 80$ to $(5b)^2 - 1$ when $b \geq 9,$ a corollary is that $(5b)^2 - 80$ is not a square, so after checking each case where $b < 9$ individually, you have shown that $(5b)^2 - 80$ cannot be a square.

Answer 2

So you have, $$ \frac{b}{4} + \frac{1}{a} = \frac{a}{20} $$

This gives you, $$ a^2 - 5ab - 20 = 0 $$

for $a$ and $b$ in $\mathbb{W}$.

You are expressing $a$ in terms of $b$ as, $$ a = \frac{5b \pm \sqrt{25b^2+80}}{2} $$

Since you want to prove/disprove that $a$ and $b$ exist by checking whether the discriminant is whole. That is, $$ \sqrt{25b^2+80} = n $$ where $n \in \mathbb{W}$. Or, $$ b = \frac{\sqrt{n^2 - 80}}{5} $$

Now, if we were to check it case by case, it is impossible as all we know is that $n \geq 9$ (for it to even be real).

There may however be other ways to prove the presence/absence of $a$ and $b$.

Answer 3

Note that $\sqrt{25b^2 - 80 }$ is a whole number if and only if $25b^2 - 80$ is a square. This gives the equation $$ 25b^2 - 80 = a^2 \implies a^2-25b^2 = 80 \implies a^2-(5b)^2 = 80$$ This can be factored as $$ (a-5b)(a+5b) = 80.$$ Using the prime factorization of $80 = 2^4 \cdot 5$ and that $a$ and $b$ are whole numbers, this results in a number of equation systems of the form $$ \begin{cases} a-5b = c_1 \\ a+5b = c_2 \end{cases} $$ where $c_1$ and $c_2$ are whole numbers such that $c_1 c_2 = 80$ (there are $20$ such systems: ten positive and ten negative).

Solving these will reduce your search to a finite number of possible solutions for $a$ and $b$, which you can sort out manually.

Answer 4

Note that $${b\over 4}+{1\over a}={a\over 20}\implies a^2=5ab+20\implies a|20 , 5|a$$so all the possible cases are $$a\in \{-20,-10,-5,5,10,20\}$$by a simple investigation, we conclude there are no integers $a,b$ such that $${b\over 4}+{1\over a}={a\over 20}$$

P.S.

Generally, all the answers to the equation $${a\over 4}+{1\over b}={c\over 20}$$are as follows$$(a,b,c)=\left(a,b,5a+{20\over b}\right)$$with any arbitrary $a\in \Bbb Z$ and $b|20$.