# An Inequality in Integers

### Statement

### Solution 1

$a^2\lt 7b^2$ so that $a^2\le 7b^2-1.$ In $\mathbb{Z}_7,\;$ $a^2\in\{0,1,2,4\},\;$ making $a^2=7b^2-1\;$ impossible. Thus, necessarily, $a^2\le 7b^2-2.$ But then, again, $a^2= 7b^2-2\;$ is also impossible such that, in fact $a^2\le 7b^2-3,\;$ or, $a\le\sqrt{7b^2-3}.\;$

Introduce function $\displaystyle f(x)=x+\frac{1}{x}\;$ which is monotone increasing for $x\ge 1.\;$ It follows that

$\displaystyle\left(\sqrt{7b^2-3}+\frac{1}{\sqrt{7b^2-3}}\right)^2\ge\left(a+\frac{1}{a}\right)^2$

which is equivalent to

$\displaystyle 7b^2-1+\frac{1}{7b^2-3}\ge\left(a+\frac{1}{a}\right)^2.$

In addition, since $b$ is a positive integer, $\displaystyle 1\gt\frac{1}{7b^2-3},$ such that $\displaystyle 7b^2\gt\left(a+\frac{1}{a}\right)^2.\;$ In other words, $\displaystyle 7\gt\left(\frac{a}{b}+\frac{1}{ab}\right)^2,\;$ i.e., $\displaystyle\frac{a}{b}+\frac{1}{ab}\lt\sqrt{7},\;$ as required.

### Solution 2

Note that modulo $7,\,$ a square may be equal only $0,\,$ $1,\,$ $2,\,$ or $4.\,$ It follows that if $a^2\lt 7b^2\,$ then also $a^2+3\le 7b^2.$

So assume $\displaystyle\frac{a}{b}\lt\sqrt{7}.\,$ We have $a^2\lt 7b^2\,$ and, hence $a^2+3\le 7b^2.\,$ This is the same as $\displaystyle\frac{a^2}{b^2}\left(1+\frac{3}{a^2}\right)\le 7,\,$ or, $\displaystyle\frac{a}{b}\sqrt{1+\frac{3}{a^2}}\le \sqrt{7}.$

Now note that

$\displaystyle\frac{a}{b}+\frac{1}{ab}=\frac{a}{b}\left(1+\frac{1}{a^2}\right)\le\frac{a}{b}\sqrt{1+\frac{3}{a^2}}\le\sqrt{7}.$

The last step was possible because $\displaystyle\frac{1}{a^2}\lt 1\,$ and $1+x\le\sqrt{1+3x},\,$ for $0\le x\le 1.$

### Solution 3

The case of $a=1\,$ is clear, so assume $a\ge 2.$

The hypothesis can be written as $\displaystyle b\gt\frac{a}{\sqrt{7}}\,$ and the conclusion as $a^2+1\lt\ ab\sqrt{7}.$

Given $a\ge 2,\,$ the conclusion is true for some $b\gt 0,\,$ if and only if it is true for

$\displaystyle b_0=\left\lceil\frac{a}{\sqrt{7}}\right\rceil.$

From $\displaystyle b_0\gt\frac{a}{\sqrt{7}},\,$ we have $7b_0^2\gt a^2,\,$ but $7b_0^2=a^2+1\,$ and $7b_0^2=a^2+2\,$ are not possible since $a^2\,(\text{mod}\,7)\,$ is $0,1,2,\,$ $4.$

Therefore, $7b_0^2\ge a^2+3\,$ so that $ab_0\sqrt{7}\ge a\sqrt{a^2+3}\gt a^2+1,\,$ as required, since $a\ge 2.$

### Illustration

### Generalization

Let $p\,$ be a prime of the form $8n+7\,$ for $n\ge 1.\,$ If $a,b\gt 0\,$ are positive integers and $\displaystyle \frac{a}{b}\lt\sqrt{p}\,$ then $\displaystyle \frac{a}{b}+\frac{1}{ab}\lt\sqrt{p}.$

The proof that is essentially the same as in Solution 3 depends on the following

**Lemma**

If $p\,$ is a prime and $p= 7\,(\text{mod }8)\,$ then neither $-1\,$ nor $-2\,$ is a quadratic residue $\text{mod }p.$

Assume $p= 7\,(\text{mod }8),\,$

Then $-1\,$ is a quadratic residue $\text{mod }p\,$ if and only if $p=1\,\text{mod }4\,$ which would entail $3=0\,(\text{mod }4),\,$ and is, therefore, not possible.

Also, $-2\,$ is a quadratic residue $\text{mod }p\,$ if and only if $p=1,3\,\text{mod }8\,$ which would entail $3=0\,(\text{mod }4),\,$ or $1=0\,(\text{mod }2),\,$ and is, therefore, not possible.

(More on this statement can be found, e.g., in T. Nagell, Introduction to Number Theory, Chelsea Publishing Co., 1981. (Sections 38,39))

**Proof of the generalization**

The hypothesis can be written as $\displaystyle b\gt\frac{a}{\sqrt{p}}\,$ and the conclusion as $a^2+1\lt\ ab\sqrt{p}.$

The case of $a=1\,$ the hypothesis is $\displaystyle b\gt\frac{1}{\sqrt{p}}\,$ and the conclusion is $2\lt\ b\sqrt{p}\,$ which is true since $b\ge 1\,$ and $p\ge 7.$

Given $a\ge 2,\,$ the conclusion is true for some $b\gt 0,\,$ if and only if it is true for

$\displaystyle b_0=\left\lceil\frac{a}{\sqrt{p}}\right\rceil.$

From $\displaystyle b_0\gt\frac{a}{\sqrt{p}},\,$ we have $pb_0^2\gt a^2,\,$ but $pb_0^2=a^2+1\,$ and $pb_0^2=a^2+2\,$ are not possible since $a^2\ne -1,-2\,(\text{mod}\,p)\,$ when $p=7\,(\text{mod}\,8).$

Therefore, $pb_0^2\ge a^2+3\,$ so that $ab_0\sqrt{p}\ge a\sqrt{a^2+3}\gt a^2+1,\,$ as required, since $a\ge 2.$

### Conjecture

It was observed that for some primes in the form $8k+5,\,$ the above statement fails. For example:

$\displaystyle \begin{cases} 5:&\frac{2}{1}\lt\sqrt{5},\,\text{but}\,\frac{2}{1}+\frac{1}{2\cdot 1}\gt\sqrt{5},\\ 13:&\frac{18}{5}\lt\sqrt{13},\,\text{but}\,\frac{18}{5}+\frac{1}{18\cdot 5}\gt\sqrt{13},\\ 37:&\frac{6}{1}\lt\sqrt{37},\,\text{but}\,\frac{6}{1}+\frac{1}{6\cdot 1}\gt\sqrt{37},\\ 61:&\frac{29718}{3805} \lt \sqrt{61},\,\text{but}\,\frac{29718}{3805} + \frac{1}{29718\cdot 3805} \gt \sqrt{61}. \end{cases}$

Given prime $p=5\,(\text{mod}\,8),\,$ are there such $\displaystyle m,n\,\text{that}\,\frac{m}{n}\lt\sqrt{p},\,\text{but}\,\frac{m}{n}+\frac{1}{m\cdot n}\gt\sqrt{p}\text{?}.$

### Acknowledgment

The above inequality, due (1979) to professor Radu Gologan, has been posted at the CutTheKnotMath facebook page by Leo Giugiuc along with a solution by Dan Sitaru and Leo Giugiuc. Radu Gologan is the Romanian team leader for the IMO. Solution 2 is by Sam Walters; Solution 3 and the beautiful illustration, the generalization and conjecture are by Gary Davis.

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