0

The positive integers $a<b$ are such that $\frac{a+b}{2}$ and $\sqrt{ab}$ are positive integers consisting of the same two digits in reverse order. What is the minimum value of a?

The first thing that comes into mind is trial and error as there are just 100 ways, but I need a fast way to do it.

I tried letting $10x + y$ and $10y + x$ be the 2 numbers

from the 2 equations,

$\frac{a+b}{2} = 10x + y$

$\sqrt{ab} = 10y +x$

$100y^2 +10xy +x^2 = 20ax +2ay-a^2$

A quadratic equation in $a$

$a^2 -2ay-20ax+100y^2+10xy+x^2$

but getting the discriminant and finding integer solutions is really tedious.

SuperMage1
  • 2,486

2 Answers2

2

$$\frac{a+b}2=10x+y \implies a^2+2ab+b^2=2^2(10x+y)^2$$

$$\sqrt{ab}=10y+x \implies 4ab=4(10y+x)^2$$

Subtracting the equations give us

$$(a-b)^2=4(9x-9y)(11x+11y)=4(9)(11)(x-y)(x+y)=4(9)(11)(x^2-y^2)$$

Hence there exists $n\in \mathbb{Z}^+$, such that $(x-y)(x+y)=11n^2$.

We know that $x \le 9$, $ 11n^2=x^2-y^2\le 81 \implies n^2 \le \frac{81}{11}<9$, hence $n^2 \in \{1,4\}$.

If $n^2=4$, then we have $$(x+y)(x-y)=44$$

Let's consider the factorization of $44$.

\begin{align} 44&=1 \times 44 \\ &=2 \times 22 \\ &= 4 \times 11 \end{align}

Since $x+y \le 18$, then we must have $x+y = 11$, and $x-y=4$. However, this would lead to $2x=15$.

Hence, we must have $n^2=1$.

$$x+y=11, x-y=1$$

That is $x=6, y=5$.

$$\frac{a+b}2=65, \sqrt{ab}=56$$

$$(a-b)^2=4(9)(11)(x^2-y^2)=(2\cdot 3 \cdot 11)^2$$

$$a-b=-66$$ $$a+b=130$$

Hence $a=\frac{64}2=32.$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
1

$$a+b=20d+2u,\\ab=(10u+d)^2,$$ implies that $$(b-a)^2=(20d+2u)^2-4(10u+d)^2=6^2\,11\,(d+u)(d-u)$$ must be a perfect square. Then $d+u$ must be $11$, while $d-u$ must be a odd perfect square (i.e. one of $1,9$, but $9$ doesn't work).

The only match is $65$, giving $a+b=130,a-b=66$ then $a=32,b=98$.