An algebra problem by Áron Bán-Szabó

Relevant wiki: Componendo and Dividendo

$\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}$

Multiplying both sides by $c+d$ and dividing by $a-b$ we get

$\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}$

By Componendo and Dividendo

$\dfrac{a}{b}=\dfrac{c}{d}$

Cross-multiplying

$ad=bc$

Hence

$abcd=(ad)^2$ must be the square of an integer

Among the choices, $\boxed{2025}$ is the only perfect square

$2025$ is in fact achievable for example when $a=c=45,b=d=1$

---

Note:

In the first step I assumed $a\neq b$ since $a=b$ in this case would imply $a=b=0$ , which means that $abcd=0$ that is not in the options

Multiply by the common denominator $(c+d)(c-d)$ : $(a+b)(c-d) = (a-b)(c+d).$ Expand the products: $ac - ad + bc - bd = ac + ad - bc - bd.$ Cancel the terms $ac$ and $bd$ ; combine the remaining terms, and divide by two. $bc = ad =: K.$ Therefore $abcd = (ad)(bc) = K^2$ must be a perfect square. This leaves only $\boxed{2025} = 45^2$ .

Example that this is indeed possible: take $a = 5$ , $b = 3$ , $c = 15$ , $d = 9$ . Then $\frac{3+5}{15+9} = \frac 8{24} = \frac 13 = \frac 2 6 = \frac{5-3}{15-9}.$

Relevant wiki: Componendo and Dividendo

$\frac{a+b}{c+d}$ = $\frac{a-b}{c-d}$

or, $\frac{a+b}{a-b}$ = $\frac{c+d}{c-d}$

or, $\frac{a+b+a-b}{a+b-a+b}$ = $\frac{c+d+c-d}{c+d-c+d}$ ............[by componendo & dividendo]

or, $\frac{a}{b}$ = $\frac{c}{d}$

or, $ad=bc$ = $k$

as they are integer $abcd$ must be a square.

there is only 1 square number 2025= $45^2$ .......[the answer]

$\frac{a+b}{c+d} = \frac{a-b}{c-d}$

By cross multiplication

$ac - ad +bc -ab = ac - bc + ad -bd$

$ad = bc$

$ad = bc$

Multiplying by $bc$ on both sides we get,

$abcd = (bc)^2$ which implies that $abcd$ must be a perfect square number So, the answer is $2025$ .

My solution was a little more practical, and less technical than some others, so I'll share it here.

• 2017 is a prime number, so it can't work.
• 2020 has prime factors 2x2x5x101, so it's a candidate (it's possible to factor it into a multiple of 4 numbers).
• 2023 has prime factors 7x17x17, not possible to factor it any further so it can't work.
• 2025 has prime factors 3x3x3x3x5x5, so also a candidate since this can also be written 9x9x5x5.

Given the equation above, intuitively, 2025 seems most likely to work so start with that:

$\frac{9+5}{9+5}$ = $\frac{9-5}{9-5}$

Since both evaluate to 1, this works. No combination of 2,2,5,101 works in this equality, so the answer is 2025. I know this isn't as rigorous as actually simplifying the equation like the other solutions here, but it does work and I like it as a more intuitive approach.

My idea is very simple. If you try to find the factors of all those answer, you will soon find out only 2025 can fit into the formula.

2017 is a prime number so no for sure.

Prime factorization of other 4 are:

• 2020: 2^2 * 5 * 101
• 2022: 2 * 3 * 337
• 2023: 7 * 17^2
• 2025: 3^4 * 5^2

Problem solved

We have: $\frac{a + b}{c + d}$ = $\frac{a - b}{c - d}$ => (a + b).(c - d) = (a - b).(c + d) => ac + bc - bd -ad = ac - bc + ad - bd => ac + bc - bd - ad - ac + bc - ad + bd = 0 => 2bc - 2ad = 0 => bc = ad => a x b x c x d = ad x bc = ad.ad But only 2025 is a square number. So E is correct.

ac+bc-bd-ad=ac+ad-bc-bd 2bc=2ad bc=ad The answer would be a number which is a perfect square. Thus, 2025. Since sqrt2025 is 45, we let, for example, a as 45 and d = 1. And if c=45, b=1, we get 1=1, after plugging in. :)

abcd must be perfect square ie2025.