This is the only title I could think of

Number Theory Level 4

How many ordered integer pairs $(a,b)$ satisfy the equation $a^{2} + 3ab - 2b^{2} = 124.$

Image Credit: Wikimedia Quadratic Equation Discriminant by KSmrq

The answer is 0.

4 solutions

Brian Charlesworth
Mar 15, 2015

Multiply through by $4$ to get

$4a^{2} + 12ab - 8b^{2} = 496 \Longrightarrow (2a + 3b)^{2} - 17b^{2} = 496$

$\Longrightarrow (2a + 3b)^{2} \equiv 3 \pmod{17}.$

But $3$ is not a quadratic residue of $17$ , so there are $\boxed{0}$ solution pairs.

Hello, How are you?

I had to guess the ans in the end :p. I found a similar problem in a book.

The problem states that "If $\cfrac { a+b }{ b } =\cfrac { b }{ a+b }$ then $a$ and $b$ must be...?" The answer is "If b is real, a is not real. If b is not real, a may be real or not real." The solution used the fact that the equation is quadratic in $a$ .

I did the same with your problem. And I got this : $\displaystyle a=\cfrac { -3b\pm \sqrt { 17{ b }^{ 2 }+496 } }{ 2 }$ & $\displaystyle b=\cfrac { -3a\pm \sqrt { 25{ a }^{ 2 }+1984 } }{ 4 }$ . I don't know how to proceed further from here.

When you get some free time, look into it. Thanks :)

Soumo Mukherjee - 6 years, 3 months ago

Hello. Good to hear from you. :)

Well, with $\frac{a + b}{b} = \frac{b}{a + b}$ we essentially have the equation $x = \frac{1}{x}$ with $x = \frac{a + b}{b}.$ This gives us $x^{2} = 1$ which has solutions $x = 1$ and $x = -1.$

In the first case we would have $a + b = b \Longrightarrow a = 0$ , in which case $b$ can be any non-zero real number.

In the second case we would have $a + b = -b \Longrightarrow a = -2b$ , which would yield an infinite number of (non-zero) solution pairs. So I'm not sure how your book came up with its solution.

As for solutions for $a$ and $b$ in your expressions, we would first observe that $17b^{2} + 496$ must be a perfect square in order for $a$ to have a chance of being an integer. This requires that $17b^{2} + 496 = m^{2}$ for some integer $m$ .

This can be written as $m^{2} = 17b^{2} + 17*29 + 3 = 17(b^{2} + 29) + 3.$ Thus we must have that $m^{2} \equiv 3 \pmod{17}.$ But $3$ is not a quadratic residue of $17$ , so there are no possible integer solutions for $a$ .

Hope this makes sense. :)

Brian Charlesworth - 6 years, 3 months ago

Feels good to hear that ^_^

Transforming $\displaystyle \cfrac { a+b }{ b } =\cfrac { b }{ a+b }$ to ${ a }^{ 2 }+ab+{ b }^{ 2 }=0$ , it found the value of a as $\displaystyle a=\cfrac { -b\pm \sqrt { { b }^{ 2 }-4{ b }^{ 2 } } }{ 2 } =\cfrac { -b\pm ib\sqrt { 3 } }{ 2 }$ . Then it concluded, "If b is real, a is not real. If b is not real, a may be real or not real."

Thanks for the explanation and the link, they shall come handy in similar situation.

Have a good day Mr. Charlesworth. :)

Soumo Mukherjee - 6 years, 3 months ago

@Soumo Mukherjee – O.k.. I think that they got the transformation wrong. Assuming that $b \ne 0$ and that $(a + b) \ne 0$ then when we cross-multiply we get the equation

$a^{2} + 2ab + b^{2} = b^{2} \Longrightarrow a(a + 2b) = 0,$

implying that either $a = 0$ or $a = -2b$ , as I found before.

As an aside, if we had the equation $\dfrac{a + b}{b} = \dfrac{b}{a - b}$ then we could transform this to $a^{2} = 2b^{2}$ , which, due to the Fundamental Theorem of Arithmetic, has no non-zero integer solutions. (We also could not have $a = b = 0$ as this would make the original equation indeterminate.)

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth –

Sorry!!

I checked, the original equation is $\displaystyle \cfrac { a+b }{ a } =\cfrac { b }{ a+b }$ . In LHS we have a in the denominator not b . Sorry for the trouble...

Soumo Mukherjee - 6 years, 3 months ago

@Soumo Mukherjee – Haha. No problem. In that case, the book is quite right: there are no real solutions, (noting that we can't have $a = b = 0$ either as that would make the original equation indeterminate). :)

Brian Charlesworth - 6 years, 3 months ago

Ben Habeahan
Mar 20, 2015

a^2 +3ba -2(b^2+62)=0 so we have D=9b^2+8(b^2+62)=17b^2+496>0 so (a,b) is 0

Bill Bell
Mar 18, 2015

A plot of 'half' the given relation shows that all of the interesting action occurs for 'a' in the set {8,...,25} and for 'b' in {4,...,-5}. Spinning through these to check for whether they satisfy the relation takes almost no time at all.

Real mathematicians will squirm when they read this, of course.

Haha. I have no idea how many "real" mathematicians there are on this site; I'm certainly not one. :) I had to do some number-crunching too in calculating quadratic residues, so I'm not sure what the real mathematicians would think of that, either.

Brian Charlesworth - 6 years, 2 months ago

I regret to inform you that you appear to be one to me.

Bill Bell - 6 years, 2 months ago

I admit that I tend to think like one, so in that respect, guilty as charged. :)

Brian Charlesworth - 6 years, 2 months ago

Murali Kancharla
Mar 16, 2015

@ Brain, how do we prove 3 is not a quadratic residue of 17 ? there is no integral value of x such that x^2=17k+3

i have done this using excel.

I'm not sure if there is a "nice" way of doing this other than going through $x = 0, 1, 2, ... ,15, 16$ and checking the remainder modulo 17 for each value of $x^{2}.$ So using excel is as good as any approach. Here is a link that has a list on quadratic residues modulo 1 through 20.

Brian Charlesworth - 6 years, 3 months ago

you can use the Quadratic reciprosity . (clearly 17 is not a square mod 3)

Pablo Torres - 6 years, 2 months ago

This is the only title I could think of

Image Credit: Wikimedia Quadratic Equation Discriminant by KSmrq

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