You have to be sure

Re-write the equation as:

$b^2-ab+a^2=0$

Now use the quadratic equation to solve for $b:$

$b=\frac{a \pm \sqrt{-3a^2}}{2}$

If $a$ is a non-zero real number, then $-3a^2$ is negative. Therefore, there are no real solutions to the equation other than $a= b=0.$

add -2ab on both sides

a^2 + b^2 - 2ab = ab-2ab

That makes (a-b)^2 = - ab

Square of a real number is always positive so there is no other solution than 0

We note that $a^2+b^2 \ge 0$ . For the equation to be true $a$ and $b$ must be both positive or negative. Since $a^2 \ge 0$ and $b^2 \ge 0$ , we can apply AM-GM inequality for $a, b >0$ as $a^2+b^2 \ge 2ab$ . Therefore, for $a, b \ne 0$ , there is no pair of reals $(a,b)$ satisfying the equation.

Deviding by a^2 we have new variable x=b/a Resulting equation x^2-x+1=0 has no real solution

Either $a$ and $b$ are both positive, or they are both negative, since $a^2+b^2$ is positive. If they are both negative, then their negations also constitute a solution. So wlog assume $a$ and $b$ are both positive.

Since $a(b-a) = b^2 > 0$ , we have $b > a$ .

But since $b(a-b) = a^2 > 0$ , we also have $a > b$ .

Hence no such solution exists.

Suppose the equation is true in the case where neither $a$ nor $b$ is zero. If and only if there is a solution for positive values $a$ and $b$ then is a solution for $-a$ and $-b ,$ so we only need to consider the case where both are positive.

Consider the right triangle with sides $a$ and $b$ and hypotenuse $c.$ Then $a^2 + b^2 = c^2.$ Because the hypotenuse is longer than the legs of a right triangle, $a < c$ and $b < c .$ So $ab < c^2$ and therefore $ab < a^2 + b^2 ,$ yet we assumed $ab = a^2 + b^2 ,$ so we have a contradiction.

By dividing by ab on both sides, we can rewrite to $\frac{a}{b}+\frac{b}{a} =1$

But, both a and b must both either be positive or negative and $\frac{a}{b} <1$ so that the sum of both fractions does not exceed 1. However, this automatically means that $\frac{b}{a} >1$ . It is therefore impossible to make the sum less than 1.

Other than $a=b=0$ , there isnt any other solution as

$ab=a^2+b^2$

$\implies 1=\dfrac{a}{b}+\dfrac{b}{a}$

Now by AM-GM we have

$\dfrac{a}{b}+\dfrac{b}{a} \geq 2$

So there arent any other possible solutions

Rewrite the equation as $a^2 - ab + b^2 = 0$ . Now complete the square: $\left( a^2 -ab + (\frac{1}{2}b)^2 \right) - (\frac{1}{2}b)^2 + b^2 = 0$ , so $(a - \frac{1}{2}b)^2 + \frac{3}{4}b^2 = 0.$ Since both terms on the left-hand side are squared, they are non-negative, so the values inside the squares must be equal to zero for equality to hold. This is true only when $a = \frac{1}{2}b$ and $b = 0$ , so (0,0) is the only solution.

Adding $2ab$ to both sides, we get:

$a^2 + 2ab + b^2 = (a+b)^2 = 3ab$

Subtracting $2ab$ from both sides, we get:

$a^2 - 2ab + b^2 = (a-b)^2 = -ab$

From this, we have that:

$(a+b)^2 = -3(a-b)^2 \\ (a+b)^2 + 3(a-b)^2 = 0$

Which holds true only if both $a + b = 0$ and $a - b = 0$ . Solving the system, we find that $(a,b) = (0,0)$ is the only solution.

We have

$ab = a^2 + b^2$

$\Rightarrow 2ab = 2(a^2 + b^2)$

$\Rightarrow a^2 + b^2 + (a-b)^2 = 0$

Since a and b are real. Therefore $a = b = 0$ by trivial inequality.

Using the equation given: ab=a^2+b^2 A=b=0

There is no indication of a negative possibility based on the equation. Shouldn't 1 be a possible real number.

For any real $a$ and $b$ the sum $a^2 + b^2 \ge 0$ which implies the product $ab \ge 0$ . Therefore $a$ and $b$ must have like signs.

Without loss of generality, let $b \ge a > 0$ which implies $b^2 \ge ab$ . Substitute $b^2 = ab - a^2$ to give $ab - a^2 \ge ab$ which simplifies to $0 \ge a^2$ . No such $a > 0$ exists therefore no pairs of real numbers $(a,b)$ satisfy the equation other than $a = b = 0$ .

We can re-write the equation as the following:

$a^2+b^2=ab \Leftrightarrow a^2 -ab +b^2=0$

And by multiplying both sides by $a+b$ we get: $(a+b)(a^2 -ab +b^2)=0 \Rightarrow a^3-b^3=0 \Rightarrow a=b$ .

Hence, there's no other solutions other that a=b=0.

if $b \neq 0$ then $ab = a^2+b^2 \implies a^2 - ab + b^2 = 0 \implies a = b(\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i) \in \mathbb{C}.$

$\therefore (a,b) = (0,0)$ is the only pair of real numbers that satisfies the above equation.

If there is an other pair , then a^2+b^2= ab>0

a^2+b^2+ab =ab+ab= 2ab

a^2+b^2-2ab = -ab

(a-b)^2= -ab

-ab<0

Contradiction

There is no pair of real number( non =0) satisfying this equation.

Suppose that a and b are sides of the right triangle, then ab would be squared hypothenuse. There cannot be such triangle because: Sine (cosine) would be a/sqrt(ab) or b/sqrt(ab), i.e. sqrt(a) or sqrt(b). So these square roots must be less or equal to 1. But these cannot be 1 because 1^2 + 1^2 is bigger than 1*1. So these must be less than 1. But product of any two numbers between 0 and 1 is always less than sum of their squares.

Case 1 $a=b,\space ab=a^2+b^2\implies a^2=2a^2\space(\text{False})\space(\forall)a\in\Bbb{R}^{*}$ Case 2 $a\ne b\text{ (let's say b > a), } ab=a^2+b^2, \text{ but } ab<b^2\implies ab<a^2+b^2$

Hence, this doesn't hold for any non-null values of a and b .

We have the equation $ab = a^{2} +b^{2}$

We know that $(a+b)^{2} = a^{2}+2ab+b^{2}$ .

We can rewrite our original equation as $ab = (a+b)^{2}+2ab$

Subtracting $2ab$ from both sides gives $-ab = (a+b)^{2}$

Square rooting both sides gives $\sqrt{a+b}$ = $\sqrt{-ab}$

Since $\sqrt{-ab}$ isn't real, we can conclude that $\sqrt{a+b}$ isn't real either and that the correct answer is that $a$ and $b$ cannot be real unless $a = b = 0$ .

Because of symmetry, we can assume a has at least as high of a magnitude as b . Now, assume that b is not zero. Then a^2 + b^2 > a^2 >= ab, so that assumption fails, meaning that b is zero. Because of symmetry, a is also zero.

Find ab = a^2 + b^2 for at least one of a and b not= 0.
Rewrite ab = a^2 + b^2 as ab = (aa + bb).
At least one of aa and bb is non-zero. Neither aa nor bb is negative. Therefore for the equation to be true ab must also be positive.
If a = b, then ab = aa, and ab = bb. Therefore ab = ab + ab. Contradiction.
Let a not= b. Then either 0 < ab < aa, or 0 < ab < bb. Therefore aa + bb > ab. Contradiction.
Therefore the equation cannot be true if either a or b not= 0

If a=b We have b²=2b² wich is impossible If a<b We have ab<b² si ab<a²+b² wich is impossible

Therefore there is a contradiction in every case

A graphical solution. Imagine an a x b rectangle, an a x a square, and a b x b square. The two squares are placed next to same length sides of the rectangle. It's clear the rectangle area is greater than that of the smaller square and smaller than that of the larger square.

Assume without loss of generality that $a \ne 0$ . If $b = 0$ , then we have $a^2 = 0$ , which implies $a = 0$ , so by contradiction, $b \ne 0$ .

For real, nonzero $a$ and $b$ , there is a unique value $x$ such that $b = ax$ . Substituting $ax$ for $b$ gives us $a^2 x = a^2 + a^2 x^2$ . Dividing by $a^2$ , we get $x = 1 + x^2$ or $x^2 - x + 1=0$ .

This quadratic has discriminant $-3$ , which means $x$ is a complex number. Because $a \ne 0$ , $b = ax$ must also be complex, which contradicts our assumption that $b$ is real. QED.

If we put:

p = a + b

q = a - b

then we have:

a = p + q

b = p - q

The equation:

$a^2 + b^2 - ab = 0$

becomes

$p^2 + 3q^2 = 0$

that, in R, it is true if and only if

p=0; q=0

so the only solution of the initial equation is:

a=0; b=0

We rewrite the equation as:

$a^{2} -ab + b^{2} =0$ .

If $a+b=0$ , the equation become $3b^{2}=0$ , or $a=b=0$ , which is the given solution.

If not, then multiply both sides by $a+b$ to get $(a+b)(a^{2}-ab+b^{2})=0$ , which is the same as $a^{3}+b{3}=0$ and $a+b=0$ , contradiction.

Thus, the only solution is $(x,y)=(0,0)$ .

a^2+b^2=c^2

ab represents the area of a rectangle with side a and b

c represents the diagonal within that shape

The question is therefore, can the area of a rectangle be equal to the area formed by squaring the diagonal line? No

Easy it is symmetrical in a & b. So any solution for a is also a solution for b. so a=b so it reduces to solve a^2 = 2a^2 and thus a=b=0 only

$ab = a^2 + b^2 \Rightarrow (a - b)^2 = -ab.$ Thus a or b must be negative, contradicting the first equation.

The equation can be rewritten as: b(a - b) = a^2

Since squares are always positive, a-b must be positive and hence a > b. But, if a - b > a; b should be greater than a to get the same result (a^2). This is contradictory and, thus, no solutions exist.

Ab = a^2 + b^2 Ab-A^2 = b^2 A-A^2/B = b 1-A/B = 1 Clearly A/B must be zero, and therefore A or B = 0. Therefore ab = 0, and a^2 + b^2 = 0. As square, real numbers must be positive, A^2 + B^2 must = 0 and so A and B must both be zero.

Let $ab=a^2+b^2; a,b\neq 0$ .

Case 1: $a=b$

$ab=a^2+b^2 \implies a^2=2a^2 \implies a= 0$ .

This is a contradiction.

Case 2: $a>b$

Since $a^2+b^2>0, ab=a^2+b^2$ iff $a,b>0$ or $a,b<0$ .

(i) If $a, b>0$ ,

$a>b\implies a^2>ab \implies a^2+b^2>ab$ .

This is a contradiction.

(ii) If $a,b<0$ ,

$a>b\implies ab<b^2 \implies ab<a^2+b^2$ .

This is again a contradiction.

Case 3: $a<b$

We will obtain contradictions in a similar manner as Case 2.

Therefore, there is no solution other than $x,y=0$

Consider $a, b \neq 0$

We have $a^2 + b^2 = ab$

Dividing both sides by $ab$ , we get

$\frac{a^2 + b^2}{ab} = \frac{ab}{ab}$

$\frac{a^2}{ab} + \frac{b^2}{ab} = 1$

$\frac{1}{b}$ + $\frac{1}{a} =1$

(Take note that this is valid, for $a, b \neq 0$ , so $ab \neq 0$ )

Now, think of two distinct unit fractions (fractions whose numerator is 1), that, when added up, equals exactly 1. When you think about it, there is absolutely no such solution.

Is there any other solution to $a^2 + b^2 = ab$ aside from $a, b = 0$ ? The answer is $\boxed{no.}$

Assuming that $ab \neq 0$ we can start looking for other solutions by dividing by $ab$ to get

$1=\frac{a}{b}+\frac{b}{a}$

or (setting $\frac{a}{b}=x$ )

$1=x+\frac{1}{x}$

Clearly $x$ must be positive - but there is no solution to this equation in the positive reals because the RHS is always greater than one.

To see this simply note that if one of the terms on the RHS is less than one, the other one must be greater than one!

Hence there are $\boxed{no}$ further solutions to the problem.

A different solution. Let, $b=ka$ , where $k$ is a real number. Then, $a^2+b^2=ab$ $(k^2+1)a^2=ka^2$ $\begin{aligned} k^2-k+1=0 &&\small \text{since } a≠0\end{aligned}$ So, we see that there is no solution of above equation. Therefore, the first equation is not valid for $a≠0$ .

Divide both sides by ab, since both a and b are different from 0. Let x:=a/b. Then x+(1/x)=1. So x^2-x+1=0. The discriminant is Delta=1-4=-3, which means there are no real solutions to this equation.

Rewrite the equation as $a^2-ab+b^2=0$

By completing the square, we can rewrite this equation as

$a^2-ab+b^2/4+3b^2/4=0\\(a-b/2)^2+3b^2/4=0$

Since squares of real numbers are always non-negative we must have $b = 0$ and $a-b/2 = 0$ which implies $a=0$ as well. $a=0$ and $b=0$ is the only pair that satisfies these conditions.