Find all positive integers a and b

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Find all positive integers a and b

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Source: IMO Shortlist 1996, N4

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orl

#1 h Aug 9, 2008, 3:15 PM • 3 Y

Y by Polynom_Efendi, Adventure10, Mango247

Find all positive integers $a$ and $b$ for which

$\left \lfloor \frac{a^2}{b} \right \rfloor + \left \lfloor \frac{b^2}{a} \right \rfloor = \left \lfloor \frac{a^2 + b^2}{ab} \right \rfloor + ab.$

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#2 h Aug 9, 2008, 3:20 PM • 2 Y

Y by Adventure10, Mango247

orl wrote:

Find all positive integers $a$ and $b$ for which
$\left \lfloor \frac {a^2}{b} \right \rfloor + \left \lfloor \frac {b^2}{a} \right \rfloor = \left \lfloor \frac {a^2 + b^2}{ab} \right \rfloor + ab.$

I only remember the result $(n,n^2+1)$ and $(n^2+1,n)$

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#3 h Sep 11, 2008, 9:00 PM • 1 Y

Y by Adventure10

" wrote:

we must have $a\neq b$ ,so WLOG assume that $a > b$ and let:

$\lfloor\frac {a^2}b\rfloor = \alpha,\lfloor\frac {b^2}a\rfloor = \beta$ .

so we must have:

$ \begin{eqnarray*}\alpha + \beta & = & \lfloor\frac {a^2 + b^2}{ab}\rfloor + ab \\ & = & \lfloor\frac {a^2 + b^2}{ab}\rfloor + \frac {a^2}b\cdot\frac {b^2}a \\ & \geq & 2 + \alpha\beta\Rightarrow a\lpha + \beta\geq 2 + \alpha\beta \\ & \Rightarrow & (\alpha - 1)(\beta - 1)\leq - 1$

but we have $\alpha\geq 1$ so $\beta = 0$ which means that $a > b^2$ .so assume that $a = b^2 + c$ in which $c$ is a natural number,so if we plug in $b^2 + c$ instead of $a$ in the given equation,we get that:

$b^3 + 2bc + \lfloor\frac {c^2}b\rfloor = \lfloor\frac {b^4 + 2b^2c + c^2 + b^2}{b^3 + bc}\rfloor + b^3 + bc$

$\Rightarrow (c - 1)b + \lfloor\frac {c^2}b\rfloor = \lfloor\frac {b^2(c + 1) + c^2}{b^3 + bc}\rfloor$ (1)

now the last equation obviously holds for $c = 1$ so $(a,b) = (n^2 + 1,n)$ is a solution,now let $c\geq 2$ ,from (1) we have:

$(c - 1)b\leq\frac {b^2(c + 1) + c^2}{b^3 + bc}$

which is equivalent to:

${ c^2(b^2 - 1) + b^2\{c(b^2 - 2) - (b^2 + 1)}\leq 0$ (2)

$\Rightarrow c(b^2 - 2) - (b^2 + 1)\leq 0$

$\Rightarrow 2(b^2 - 2) - (b^2 + 1)\leq 0$

$\Rightarrow b = 1\textrm{ or }2$

for $b = 1$ we get $a = 2$ ,also for $b = 2$ ,using (2) we get that:

$3c^2 + 4(2c - 5)\leq 0$

so we must have $c < 2$ ,contradiction.

so the only solutions are $(a,b) = (n,n^2 + 1)$ and $(a,b) = (n^2 + 1,n)$ in which $n$ runs over the naturals.

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tc1729

#4 h May 1, 2012, 5:02 AM • 2 Y

Y by Adventure10, Mango247

We claim that all ordered pairs of integers $(a, b)$ are of the form $(k, k^2+1)$ or $(k^2+1, k)$ for $k\in\mathbb{Z}^{+}$ .

Let $a=b^2+1$ . If $b>1$ , then we have $\text{LHS}=\left[b^3+2b+\frac{1}{b}\right]+\left[\frac{b^2}{b^2+1}\right]=b^3+2b,$ and $\text{RHS}=\left[b+\frac{1}{b}+\frac{b}{b^2+1}\right]+b^3+b=b^3+2b,$ so we have a solution. It is easy to confirm that $b=1$ also yields a solution. There is an obvious symmetry between $a$ and $b$ , so $b=a^2+1$ is also a solution.

Now we must show there are no other solutions. WLOG, asume $b\ge a$ .

Let $a$ be fixed and let $f(b)=\frac{a^2}{b}+\frac{b^2}{a}-\left(\frac{a}{b}+\frac{b}{a}\right)-ab$ and $g(b)=\frac{a^2}{b}+\frac{b^2}{a}-\frac{a}{b}-\frac{b}{a}-ab$ . From this we have $|g(a)-f(a)|<3$ .

Observe that we have $g(a)=\frac{1}{a}\left(b-\frac{a^2+1}{2}\right)^2=\frac{1}{4m}\left(\frac{a^2+1}{2}\right)^2=\frac{a^2-a}{b}.$ So if $b\ge a^2+2$ , then $g(b)\ge a+\frac{2}{a}+\frac{a^2-a}{a^2+2}>a.$ Hence $f(b)>a-3$ . So for $m>2$ there are no solutions for $b>a^2+1$ .

Also if $a\le b\le a^2-1$ , then $\left(b-\frac{a^2+1}{2}\right)^2\le\frac{(a^2-3)^2}{4}$ and $\frac{a^2-a}{b}\le a-1$ , so $g(b)\le -a+\frac{2}{a}-1<0$ . Hence $f(b)<0$ for $b>2$ . So for $b>2$ there are also no solutions for $b<a^2$ . It is also easy to check that $f(a^2)=-(a-1)<0$ .

Now, it remains to check $a=1$ and $a=2$ . For $a=1$ , we have $f(1)=-1$ . If $b>1$ then we have $f(b)=0+b^2-2b=b(b-2)$ . So for $a=1$ the only solution is $b=2$ .

Finally take $a=2$ , and consider $b\ge 2$ . It is easy to check that $f(2)=-2$ , $f(3)=-3$ , and $f(4)=-1$ . For $b\ge 6$ we have $f(b)=0+\frac{b^2}{2}-\frac{b}{2}-2b\ge\frac{b^2}{2}-\frac{1}{2}-\frac{5b}{2}>0$ .

Hence, $(a, b)= (k, k^2+1)$ or $(k^2+1, k)$ for $k\in\mathbb{Z}^{+}$ . $\Box$

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#5 h Apr 19, 2025, 10:16 PM • 1 Y

Y by Ravensrule8

Without loss of generality $a\geq b$ . If $a=b$ , then $(a-1)^2 + 1 = 0$ , impossible, so $a>b$ .

The left-hand side does not exceed $\frac{a^2}{b} + \frac{b^2}{a}$ , while the right-hand one is at least $ab+2$ . Hence $a^3 + b^3 \geq a^2b^2 + 2ab$ , equivalent to $(a^2 - b)(b^2 - a) + ab \leq 0$ . Since $a^2 \geq a > b$ , the latter is false if $b^2 \geq a$ . Therefore we have obtained $a > b^2$ , in particular $\left \lfloor \frac{b^2}{a} \right \rfloor = 0$ .

Write $a = b^2 + c$ for some positive integer $c$ . The equation becomes $b^3 + 2bc + \left\lfloor \frac{c^2}{b} \right \rfloor = \left \lfloor \frac{2b^4 + 2b^2c + c^2}{b(b^2+c)} \right \rfloor + b^3 + bc$ , i.e. $bc + \left\lfloor \frac{c^2}{b} \right \rfloor = \left \lfloor \frac{2b^4 + 2b^2c + c^2}{b(b^2+c)}\right\rfloor$ . Hence $bc \cdot b(b^2 + c) \leq 2b^4 + 2b^2c + c^2$ , so $(b^2 - 1)c^2 + b^2(b^2 - 2)c - 2b^4 \leq 0$ . Since $b \geq 1$ , we deduce $b^2(b^2 - 2)c - 2b^4 \leq 0$ , i.e. $c(b^2 - 2) \leq 2b^2$ . If $c\geq 3$ , then $b^2 \leq \frac{2}{c-2} \leq 1$ , so $b=1$ and putting this in the main equation directly gives $c=1$ , i.e. $a=2$ . If $c=2$ , then the main equation becomes $2b + \left \lfloor \frac{4}{b} \right\rfloor = b + \left \lfloor \frac{2}{b} + \frac{b}{b^2 + 2} \right\rfloor$ , so $b + \left \lfloor \frac{4}{b} \right\rfloor = \left \lfloor \frac{2}{b} + \frac{b}{b^2 + 2} \right\rfloor$ . This is false for $b=1$ and $b=2$ by direct check, as well as for $b\geq 3$ since $\frac{2}{b} + \frac{b}{b^2 + 2} < \frac{2}{b} + \frac{1}{b} \leq 1$ . Finally, if $c=1$ , then the main equation is satisfied for any $b$ .

Therefore all solutions are $(n^2 + 1, n)$ and $(n, n^2 + 1)$ , where $n$ is an arbitrary positive integer.

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