A number theory problem by Hrishikesh Harish

Number Theory Level 2

Positive integers $a$ and $b$ are such that $a + b = \dfrac ab + \dfrac ba$ . What is the value of $a^2 + b^2$ ?

The answer is 2.

4 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 $a^2+b^2=1+1=\boxed2$

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

Why it has to be true? Maybe there are other solution that also satisfy this condition?

Pi Han Goh - 3 years, 11 months ago

Otherwise, the right-hand side would not add up to an integer, but he left would.

Siva Budaraju - 3 years, 11 months ago

It is conceivable that the right side would be two fractions which will add up to an integer, but that cannot happen either. Say $\frac ab+\frac ba$ simplifies to $\frac cd+\frac dc$ , where $c$ and $d$ do not have a common divisor other than $1$ . Then $\frac cd+\frac dc=\frac{c^2+d^2}{cd}$ . For this to be an integer, the numerator has to be divisible by $c$ . $c^2$ is divisible by $c$ , however $d^2$ is not, so the entire numerator is not divisible by $c$ and the fraction cannot be equal to an integer. From this follows that $a$ and $b$ must divide each other.

Marta Reece - 3 years, 11 months ago

@Marta Reece –

but that cannot happen either.

Why not?

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – Because of all the sentences which follow this statement in my previous comment.

Marta Reece - 3 years, 11 months ago

Really? Are you saying that solutions like a/b = 2.5, b/a = 0.5 are not possible?

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – Yes, because a/b has to be the reciprocal of b/a.

Siva Budaraju - 3 years, 11 months ago

Hana Wehbi
Jun 27, 2017

The only solution that works here is when $a=b=1 \implies a^2+b^2=2$

How do you know that's the only solution?

Pi Han Goh - 3 years, 11 months ago

If $\frac {a}{b}+ \frac{b}{a} = a+ b\implies a=b$ , where $a$ and $b$ are positive integers and if the sum of their squares is equal to $2\implies a=1=b$ .

Hana Wehbi - 3 years, 11 months ago

Why must a=b be true? How do you know that there isn't a solution for "a not equal to b"?

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – If we look at the equation $\frac{a}{b}+ \frac{b}{a}= \frac{a^2+b^2}{ab}= a+b \implies a^2+b^2= ab(a+b)$ .

We know that $a^2+b^2= 2\implies ab(a+b)= 2$ . Also, we know that $a$ and $b$ are positive integers so $ab(a+b)= 1\times 2$ .

Case1: if $a\ne b \implies ab=2 \text { and} \ a+b=1\implies \frac{a}{b}+\frac{b}{a} \ne a+b$ . This case is rejected because it contradicts our result.

Case2: if $a=b \implies ab=1 \text { and}\ a+b=2 \implies a=b=1$

Hana Wehbi - 3 years, 11 months ago

@Hana Wehbi – How do you know that a^2 + b^2= 2 in the first place?

Pi Han Goh - 3 years, 11 months ago

Alejandro Castillo
Feb 16, 2018

We have that $(a+b)ab=a^2+b^2$ then $a^2b+b^2a=a^2+b^2$ . Considere $a^2b-a^2=b^2-b^2a \implies a^2(b-1)=b^2(1-a)$ . If $b>1$ then $b^2(1-a)$ should be positive, then $1-a$ is positive, and it happen if $1>a$ but it can be because $a$ should be integer and positive. Therefore $b\le1$ and it implies that $b=1$ then $a^2(1-1)=b^2(1-a)=1(1-a)=1-a=0 \implies a=1 \therefore a=b=1 \implies a^2+b^2=2$

Aman Kumar
Jul 3, 2017

Let a/b=t for some non zero t(you can check that t has to be integer) Then the equation becomes: b=(t^2 +1)/(t^2+t) Which in turn gives a=(t^2+1)/(t+1). As b is an integer =>(t^2+1)=(t^2+t)k for some integer k Which is possible only for t=1... Hence a=b=1 Thus, a^2+b^2=2

A number theory problem by Hrishikesh Harish

The answer is 2.

4 solutions

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