An algebra problem by sam dave

Algebra Level 2

Given that $a$ and $b$ are two positive integers such that $a + b = \dfrac{a}{b} + \dfrac{b}{a}$ , what is the value of $a^{2} + b^{2}$ ?

2.5 2 1 4 3.5

2 solutions

Marta Reece
Jun 28, 2017

$\frac ab+\frac ba$ has to be an integer, since it is equal to an integer $a+b$ .

So $a$ has to be divisible by $b$ and $b$ in turn has to be divisible by $a$ .

This can only happen if $a=b$ .

The equation $a+b=\frac ab+\frac ba$ then becomes $2a=1+1$ and has only one solution $a=1$

So $a=b=1$ and $a^2+b^2=1+1=\boxed2$

$\frac ab+\frac ba$ has to be an integer, since it is equal to an integer $a+b$ .

So $a$ has to be divisible by $b$ and $b$ in turn has to be divisible by $a$ .

Why must the latter equation be true? Why can't we have (a/b) and (b/a) to be fractions, but they still adds up to an integer?

Pi Han Goh - 3 years, 11 months ago

Let's assume $\frac ab+\frac ba=\frac cd+\frac dc$ where $\frac cd$ is a simplified fraction. Then $\frac cd+\frac dc=\frac{c^2+d^2}{cd}$ . For this to be an integer, the numerator must be divisible by $c$ . $c^2$ is divisible by $c$ , however $d$ is not. So the entire numerator is not divisible by $c$ and the fraction cannot be simplified to an integer. Similarly for $d$ . So $\frac ab+\frac ba$ can be an integer only if $a$ and $b$ divide each other as stated.

Marta Reece - 3 years, 11 months ago

What do you mean by c/d is a simplified fraction? Why can we assume that?

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – By simplified fraction I mean that there is not common divisor of $c$ and $d$ other than $1$ . And yes, we can assume that. Either $\frac ab$ is already simplified, or there is $\frac cd$ as described, or $a$ and $b$ divide each other, and therefore the situation does not arise in the first place.

Marta Reece - 3 years, 11 months ago

given that a/b + b/a = a+b,now this can be rewritten as a( 1/b -1) + b(1/a-1)=0 now as a and b are positive integers they cant be zero so 1/b-1 and 1/a-1 has to be zero to get overall result as zero, so 1/b-1=1/a-1=0 that gives a=b=1,hence value of a squared and b squared is equal to 1+1 = 2

sam dave - 3 years, 11 months ago

Why must 1/b - 1 = 1/a - 1 = 0? How do you know that there's no solution to "a(1/b - 1) = -b(1/a - 1) = k" for some non-zero k?

Pi Han Goh - 3 years, 11 months ago

it is given that ,a and b are positive integers so they cant be zero,and we need the sum of that expression as zero so1/b-1=1/a-1=0 ,has to be the only possible case , as sum of two positive quantities cannot be zero,even if you consider a(1/b - 1) = -b(1/a - 1) = k" for some non-zero k ,they k has to be zero

sam dave - 3 years, 11 months ago

Sam Dave
Jun 26, 2017

HINT: use the fact and a and b are positive intgers . if the solution is needed ,let me know. given that a/b + b/a = a+b,now this can be rewritten as a( 1/b -1) + b(1/a-1)=0 now as a and b are positive integers they cant be zero so 1/b-1 and 1/a-1 has to be zero to get overall result as zero, so 1/b-1=1/a-1=0 that gives a=b=1,hence value of a squared and b squared is equal to 1+1 = 2 – Sam Dave · now

Please add the solution. Thanks.

Since $a$ and $b$ are positive integers, having decimal options doesn't really make sense.

Calvin Lin Staff - 3 years, 11 months ago

An algebra problem by sam dave

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