Question

Two positive integers a and b are such that a + b = (a/b) + (b/a). What is the value of a² + b²?
Provide detailed solution...
10 points.....

Answer 1

Answer:

2

Step-by-step explanation:

a+b=a/b+b/a

-> Now take L.C.M

a+b=(a2+b2)/ab

-> Cross multiplying

(a+b)ab=a2+b2

a2.b+b2.a=a2+b2

Take a2 and b2 common

a2(b−1)+b2(a−1)=0 --(1)

Now since a and b are positive integers - their square can’t be zero.

So in order to make the equation 1 equal to zero:

(b−1) and (a−1)both has to be 0.

Therefore,

b−1=0 =>b=1

and,

a−1=0 =>a=1

Hence a=b=1

Thus a²+b²=1²+1²

                   =2

Answer 2

Answer:

The value of  a² + b² is 2.

Step-by-step explanation:

Integers are the group of whole numbers and negative numbers. Identical to whole numbers, integers also do not contain the fractional part. Therefore, we can say, integers exist as numbers that can exist as positive, negative, or zero, but cannot be a fraction.

Given,

a + b = (a/b) + (b/a)

To find,

value of a² + b²

Step 1

$&a+b=a / b+b / a \Rightarrow a+b=\frac{a^{2}+b^{2}}{a b}$

$&a^{2} b+a b^{2}=a^{2}+b^{2}$

$&a^{2}(b-1)+b^{2}(a-1)=0$

because $\mathrm{a}, \mathrm{b}$ are positive integers, $\mathrm{a}^{2}, \mathrm{~b}^{2}$can't be zero

$\Rightarrow \mathrm{a}-1=0 \Rightarrow \mathrm{a}\\=1\mathrm{~b}-1=0 \Rightarrow \mathrm{b}=1$

$a^{2}+b^{2}=1+1=2$

Hence, The value of  a² + b² is 2.

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