Inspired by Marta Reece 2

For a simple counter-example, take $A=\sqrt{2}, B=-\sqrt{2}$ .

Given

$a^2+b^2=x$
$ab=y$
$a+b=z$

where $x, y, z$ are integers. The difference $a-b$ then satisfies

$(a-b)^2 = x-2y$

which is an integer. However, the square root of an integer isn't necessarily even rational. Let's say

$a-b = w$

where $w$ is an irrational number. Then

$(a+b)+(a-b) = 2a = z + w$

which is irrational if $w$ is. So, it's possible that neither $a$ or $b$ is an integer. As an example, the following values for $a,b$ will yield integer $x,y,z$

$a= \dfrac{1}{2}(3+\sqrt{5})$

$b= \dfrac{1}{2}(3-\sqrt{5})$

Relevant wiki: Quadratic Equations - Problem Solving

First note that if $AB$ and $A+B$ are integers, then $A^2 + B^2 = (A+B)^2 - 2(AB)$ must be an integer, so we can simplify the assumption to simply that $AB$ and $A+B$ are integers.

Now consider the quadratic with roots $A$ and $B$ . It factors as $(x-A)(x-B) = x^2 - (A+B)x + AB,$ but since every quadratic has this property all we know about $A$ and $B$ is that they are the roots of a quadratic polynomial with integer coefficients (and leading coefficient $1$ ).

Now we can rephrase the question in a way that every student of algebra should be able to answer:

"Must every quadratic with leading coefficient $1$ and integer coefficients ( $A+B$ and $AB$ ) have integer roots ( $A$ and $B$ )?"

or another way:

"Can every quadratic with integer coefficients be factored using integers?"

To both, the obvious answer is "no".

For instance, $x^2 - 2 = (x-\sqrt{2})(x+\sqrt{2}) \\ x^2 - x - 1 = (x-\phi)(x-\theta)$ both give counterexamples.

Last, to give a necessary and sufficient condition for all real counterexamples, we note by the quadratic formula that $A$ and $B$ are $A,B = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$ for integers $b,c$ such that $b^2 - 4c > 0$ is not a perfect square.

In other words, the complete set of real counterexamples has the form $A,B = \frac{m}{2} \pm \frac{\sqrt{n}}{2}$ where $m,n$ may be freely chosen integers such that $n > 0$ is not a perfect square and $m^2 \equiv n \pmod{4}$ .

Over the real numbers, any conjugate pairs of the form $A = a + b \sqrt{c}$ and $B = a - b \sqrt{c}$ , where $a,b,c \in \mathbb{Z}$ are possible for $A$ and $B$ .

• $A + B = (a + b \sqrt{c}) + (a - b \sqrt{c}) = 2a \in \mathbb{Z}$

• $AB = (a + b \sqrt{c})(a - b \sqrt{c}) = a^2 - b^2 (\sqrt{c})^2 = a^2 - b^2c\in \mathbb{Z}$

• $A^2 + B^2 = (a + b \sqrt{c})^2 + (a - b \sqrt{c})^2 = (a^2 + 2ab \sqrt{c} + b^2c) + (a^2 - 2ab \sqrt{c} + b^2c) = 2a^2 + 2b^2c\in \mathbb{Z}$

Similarly, if complex solutions are permitted, then any conjugate pairs of the form $A = a + b \sqrt{c}\ i$ and $B = a - b \sqrt{c}\ i$ , where $a,b,c \in \mathbb{Z}$ are possible for $A$ and $B$ .

Counter Example

$A=i\\ B=-i\\ A+B=0\\ AB=1\\ { A }^{ 2 }+{ B }^{ 2 }=-2$

The counter example I was thinking of was $A=i$ and $B=-i$ .

Let A equal some square root that isn’t equivalent to an integer and B=-A

Then if you do A^2+B^2, the output is 2A^2, which is an integer since A^2 is an integer

AB=A(-A)=-A^2 which is an integer for the same reason

A+B=0 since B=-A.

First, realize that $(A+B)^2-2AB=A^2+B^2$ . This means that if the sum is an integer and the product is an integer, the sum of the squares is guaranteed to be an integer. Then, consider Vieta sums for a quadratic function that has A and B as roots. The product is the constant term, and the sum is the (negated) first-degree term. Since we know there are quadratics that have integer coefficients but non-integer roots, we can conclude that it is possible for A and B to be non-integers.

By squaring A+B and subtracting 2(AB) from it, we obtain A^2+B^2 which should also be an integer. So, we can pick any 2 integers for A+B and AB to get any integer A^2+B^2. A+B=AB=1 would suffice to show that A and B doesn’t need to be integers.

So you found a counterexample... Now, can you give a general description of the numbers $A,B$ that satisfy the description?

Here is the follow-up question!

This problem just requires a counterexample.

Let $\mathbb{Z}$ be the set of all integers and choose $A = \sqrt{2}$ and $B = -\sqrt{2} \implies A + B = 0 \in \mathbb{Z},$ $AB = -2 \in \mathbb{Z}$ and $A^2 + B^2 = 4 \in \mathbb{Z}$

$\therefore A^2 + B^2 \in \mathbb{Z}, AB \in \mathbb{Z}$ and $A + B \in \mathbb{Z}$ $\nRightarrow$ $A \in \mathbb{Z} \wedge B \in \mathbb{Z}$ .

Assume A,B are in the form of x+√y and x-√y (x,y are integers)

Take A as $i$ and B as $-i$ .( $i^2$ =-1)

If we consider a=i^2 b= -i^2 then a^2+b^2 is an integer , ab is an integer , a+b is also an integer . However a, b are not integers.

For a “complex” counterexample, let $A = i$ and $B = -i$ , so $A^2 + B^2 = i^2 + (-i)^2 = -2$ , $AB = (i)(-i) = 1$ , and $A + B = i + (-i) =0$

A and B can be any number of the form $a-\sqrt{b}$ and $a+\sqrt{b}$ where a and b are integers

A=1+i, B=1-i.

Seems like this should be an easy problem

(a-b) 2=(a+b) 2-4ab=integer. a-b may be non-integer. So are a and b.

A = i B = 1/i

No. (and the same question could be asked with "real" in place of "integer", and probably has elsewhere.)

A simple counter example is $A = \sqrt {2}$ and $B = - \sqrt {2}$ when $A^2 + B^2$ , $AB$ , and $A + B$ would be 4, -2, and 0 respectively.

The analogous question for reals using i and -i works similarly.

$A = i$ and $B = - i$ when $A^2 + B^2$ , $AB$ , and $A + B$ would be 0, 1, and 0 respectively, all real..

Suppose that $I) A+B=N$ and $II) AB=M$ with N and M being two integers.

Then it automatically follows that $A^2+B^2$ is also an integer because $A^2+B^2=(A+B)^2-2AB$ . It means that we don't have to care about the third condition.

Now replace $A=\frac{M}{B}$ in $I)$ . After some manipulation, we get the quadratic equation $B^2-NB+M=0$

Solving the quadratic equation gives us $B_1 =$ $\frac{\sqrt{(N^2-4M)}+N}{2}$ , $B_2 =$ $\frac{-\sqrt{(N^2-4M)}+N}{2}$ . Because $B_1 B_2 =M$ , we can simply say that $B_1=A$ and $B_2=B$

Now, we only have to input values for N and M such that the value under the square root is not a square number (and not negative) then A and B won't be integers.

A nice counter-example: If we choose $N=1$ and $M=-1$ we get both golden ratios for A and B with $A=\frac{1+\sqrt{5}}{2}$ and $B=\frac{1-\sqrt{5}}{2}$

Hmm... Actually I think $i$ and $-i$ isn't integer. So let $A=i$ , $B=-i$ , then $A^2+B^2=-1-1=-2$ , $AB=i^2 \times -1=1$ , $A+B=-i+i=0$ . All is integer.

A = -sqrt(B) or B = -sqrt(A)

Even if A & B are conjugate complex numbers the above statement holds good let 'z=x+iy' denote a non-real complex number and let 'z =x-iy' be its conjugate,where x & y are integers numbers then z+z and zxz* are also integers and even square of z & z* are conjugate with respective of each other:Thus same properties are observed Hence proved.

Let $A=i$ and $B=-i$

Draw a square where the side is A+B, then there are infinite solutions of (A,B) pairs along the diagonal of the square

Let A=i and B=-i then ${ A }^{ 2 }+{ B }^{ 2 }={ i }^{ 2 }+{ (-i) }^{ 2 }=-2$ , ${ A }{ B }={ i }{ (-i) }=1$ and ${ A }{ +B }={ i }{ +(-i) }=0$

Consider A=I1 +f ,B=I2 -f and Now solve (A+B)^2 = A^2 +B^2 +2AB Put as above considered you will get f(I2 -I1) =0 FOR I1 =I2 ,IT IS POSSIBLE FOR ALL VALUES OF FRACTIONAL PART f Hence A and B not necessiraly have to be integers

Wow everyone's explanations are awesome and complex. Here all I was thinking A + B has to be an integer is false thus No.

A simple counterexample: $A=\sqrt{1}; B=\sqrt{4}$

A counterexample:
A=-√２,B=√２. A²+B²=0 AB=-2 A+B=0

These 3 facts will become useful when finding a solution to this problem:

• A non-real number cannot be an integer (but can be a Gaussian integer).

• A Gaussian integer, $x$ , times another Gaussian integer, $y$ , where the imaginary part of $y$ is -1 times the imaginary part of $x$ is only an integer if the real parts of $x$ and $y$ are equal (if they are complex conjugates).

• A Gaussian integer squared will only be an integer if it is purely real or purely imaginary.

Now, using those facts, let's try finding a solution where $A$ is non-real.

First, because $AB$ is an integer, if $A$ is non-real, then $B$ must also be non-real (as a non-real number times a purely real number cannot equal another purely real number).

Next, because $A+B$ is an integer, and because $A$ and $B$ are both non-real, the imaginary part of $B$ must be -1 times the imaginary part of $A$ . Because of the second fact listed above, this means that $B$ must be the complex conjugate of $A$ .

Finally, because ${ A }^{ 2 }+{ B }^{ 2 }$ is an integer if ${ A }^{ 2 }$ and ${ B }^{ 2 }$ are both integers, $A$ and $B$ could either be both purely real or both purely imaginary (because of the third fact listed above). Because we want to find a solution where either $A$ or $B$ is imaginary, and because $B$ is the complex conjugate of $A$ , $A$ and $B$ must both be purely imaginary.

We now have our answer: If A and B are both purely imaginary, if they are both integers upon being divided by $i$ , and if $A=-B$ , then ${ A }^{ 2 }+{ B }^{ 2 }$ , $AB$ , and $A+B$ must all be integers without either $A$ or $B$ being an integer.

Footnote: This is only one way of solving this problem. There are other valid ways to solve this problem that don't involve imaginary numbers.