Pre RMO 2015- inequality or pure algebra?

Algebra Level 2

Positive integers $a$ and $b$ are such that $a + b = \frac{a}{b} + \frac{b}{a}$ . What is the value of $a^{2} + b^{2}$ ?

The answer is 2.

6 solutions

Brian Charlesworth
Oct 28, 2015

We can rewrite the given equation as

$a + b = \dfrac{a^{2} + b^{2}}{ab} \Longrightarrow a^{2}b + ab^{2} = a^{2} + b^{2} \Longrightarrow a^{2}(b - 1) = b^{2}(1 - a).$

Then since $a^{2}, b^{2} \gt 0$ and $a,b$ are positive integers we see that the two sides of this last equation will have the same sign if and only if $a = b = 1,$ (in which case they are both zero). As these values do indeed satisfy the original equation we can conclude that these solution values are unique and that $a^{2} + b^{2} = \boxed{2}.$

How do you start your approach when you see such a question?

Debmeet Banerjee - 5 years, 7 months ago

Best solution here! Upvoted.

Pi Han Goh - 5 years, 7 months ago

By "parity", you mean sign (negative or non-negative), right? Because the usual definition of parity (even or odd) doesn't seem to be applicable here.

Prasun Biswas - 5 years, 7 months ago

Yes, you're right. Using parity in this context is a bad habit I picked up long ago which resurfaces now and then. I've made the appropriate edits. Thanks for pointing that out. :)

Brian Charlesworth - 5 years, 7 months ago

Now that it has been edited accordingly, repeating what Pi Han already said above, this is the best solution here! +1 :)

Prasun Biswas - 5 years, 7 months ago

Good! This is my approach that uses the basics of the AM - GM Inequality . So here it is:

Like what I've said, by the AM - GM Inequality , we know that $\frac{\frac{a}{b} + \frac{b}{a}}{2}$ $\ge$ $\sqrt{\frac{a}{b} × \frac{b}{a}} = 1$ or $\frac{a}{b} + \frac{b}{a}$ $\ge 2$ and the equality occurs iff $\frac{a}{b}$ = $\frac{b}{a}$ . From my last statement, we'll get that $a$ = $b$ . So from here, we substitute these values ( $\frac{a}{b} + \frac{b}{a} = 2$ and $a$ = $b$ ) into the given equation: $a$ + $b$ = $\frac{a}{b} + \frac{b}{a}$ to get $2b = 2$ , which immediately tells us that $a = 1$ and $b = 1$ . And ${a}^2 + {b}^2$ should be equal to $\boxed{2}$ .

Reineir Duran - 5 years, 5 months ago

Great approach! :)

Brian Charlesworth - 5 years, 5 months ago

Frankly speaking, I don't understand why you're using the equality case for $\frac ab+\frac ba\geq 2$ here. Nowhere in the problem are we given that $\frac ab+\frac ba=2$ which would justify you using the equality case to get $a=b$ . We're just given that $a,b$ are positive integers such that $a+b=\frac ab+\frac ba$ .

Also, in your approach, you never use the result that $a,b$ are positive integers , just that $a,b$ are positive reals. That seems faulty to me since if we just let $a,b$ be positive reals, there are many solutions other than $a=b=1$ , one of them being $(a,b)=\bigg(5,\frac 58(\sqrt{41}-5)\bigg)$

Prasun Biswas - 5 years, 5 months ago

Abdur Rehman Zahid
Oct 23, 2015

$\begin{aligned}a+b=\frac{a}{b}+\frac{b}{a}\\ a+b=\frac{a^2+b^2}{ab}\\a+b=\frac{(a+b)^2-2ab}{ab} \end{aligned}$ Let $a+b=x$ .Then the equation becomes: $\begin{aligned}x=\frac{x^2-2ab}{ab}\\ abx=x^2-2ab\\x^2-abx-2ab=0\end{aligned}$ Applying the Quadratic Formula gives: $x=\frac{ab\pm\sqrt{(ab)^2+8ab}}{2}$ $a,b\in \mathbb{Z^+}\rightarrow x\in\mathbb{Z^+}$ ,so $(ab)^2+8ab$ must be a perfect square.Hence, $(ab)^2+8ab=z^2\\ (ab+4)^2=z^2+16\\ (ab+4)^2-z^2=16\\ (ab+4+z)(ab+4-z)=16$ Solving all the possible system of equations for $ab$ and $z$ ,we get that $ab=1,z=\pm3$ .Putting value of $ab$ in the formula,we get: $x=\frac{1\pm\sqrt{1+8}}{2}\rightarrow x=\frac{1+3}{2}\rightarrow x=2$ We discard $x=\frac{1-3}{2}$ since $x\in \mathbb{Z^+}$ Hence $x^2-2=2^2-2=\boxed{2}$

BTW,this is the 2015th problem which I solved :)

Moderator note:

Given that $a, b$ are positive integers, there are much faster approaches we can take. For example, $a \geq \frac{a}{b}$ and $b \geq \frac{b}{a}$ and so we can immediately conclude that $a = b = 1$ .

2015th? Congratz! Now aim for 5102. Hahah

Pi Han Goh - 5 years, 7 months ago

Goes off to solve 3087 more problems...

Abdur Rehman Zahid - 5 years, 7 months ago

I've solved 72 more problems than that!

Sharky Kesa - 5 years, 7 months ago

Another approach: The given equation can be rewritten as

$a + b = \dfrac{a^{2} + b^{2}}{ab} \Longrightarrow a^{2}b + ab^{2} = a^{2} + b^{2} \Longrightarrow a^{2}(b - 1) = b^{2}(1 - a).$

Then since $a^{2}, b^{2} \gt 0$ and $a,b$ are positive integers we see that the two sides of this last equation will have the same parity if and only if $a = b = 1.$ As these values do indeed satisfy the original equation we can conclude that these solution values are unique and that $a^{2} + b^{2} = \boxed{2}.$

Brian Charlesworth - 5 years, 7 months ago

Thanks sir for giving the solution without using inequalities.

Priyanshu Mishra - 5 years, 7 months ago

You should post that as your solution.

Pi Han Goh - 5 years, 7 months ago

O.k.. Done. :)

Brian Charlesworth - 5 years, 7 months ago

Anand Chitrao
Oct 27, 2015

If a and b are intgers, then a/b and b/a must be integers(they are reciprocals of each other). Hence they are equal and then immediately a = b = 1 follows

It's not obvious that $\frac{a}{b}$ and $\frac{b}{a}$ are both integers. It's possible that neither is an integer, but they do add up to $a+b$ .

Ivan Koswara - 5 years, 7 months ago

Lion Soul
Oct 24, 2015

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

Taking LCM.

a+b = (a^2 +b^2)/ab

(a^2)×b+(b^2)×a = (a^2)+(b^2)

Equating coefficients. a=1 and b=1.

(a^2)+(b^2) = 2

You cannot "Equate coefficients" unless we have an algebraic identity of variables. However, all that we have is an expression that holds at certain constant values to be found.

Are you saying that if $2x + 3 y = 2 \times 6 + 3 \times 4$ then the only solution is $x = 6, y = 4$ ?

Calvin Lin Staff - 5 years, 7 months ago

Im not that good at math. I used LCM and cross multiplication to solve this question and got right answer. I cant argue with you because Im still a beginner in ALGEBRA.

lion soul - 5 years, 7 months ago

Keep on improving!

Calvin Lin Staff - 5 years, 7 months ago

@Calvin Lin – THANK YOU.

lion soul - 5 years, 7 months ago

Sir, is the problem not so less rated? I think it should be 40-50 points, more than 30 but not 25.

Priyanshu Mishra - 5 years, 7 months ago

Engki Mai Putra
Nov 17, 2015

a , b is natural number, it make LHS and RHS has same parity , because LHS natural number, it make in RHS a/b and b/a should be natural implies a=b , put to equation se get , 2a=2 if and only if a=b=1 , finally we get a^2 +b^2=2 , is my ways true?

Xiaoying Qin
Oct 28, 2015

Never said they had to be different, so a =1 and b = 1 so 1/1 + 1/1 will be 2.

Pre RMO 2015- inequality or pure algebra?

The answer is 2.

6 solutions

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