The number of integer solutions of equation $2|x|\left(x^{2}+1\right)=5 x^{2}$ is

We need to find how many integer values of $x$ satisfy the equation $2|x|(x^2 + 1) = 5x^2$.

The presence of $|x|$ (absolute value) makes this equation tricky because we need to handle positive and negative values of $x$ carefully.

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Let's use a clever substitution to simplify our work:

Set $y = |x|$

This means:

$y \geq 0$ (since absolute value is always non-negative)

$y^2 = x^2$ (since $|x|^2 = x^2$ for any real number $x$)

Our equation becomes:

$2y(y^2 + 1) = 5y^2$

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When $x = 0$:

If $x = 0$, then $y = |0| = 0$

Substituting into our equation:

Left side: $2(0)(0^2 + 1) = 0$

Right side: $5(0)^2 = 0$

Since $0 = 0$, $x = 0$ is a solution.

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When $x = 0$:

If $x= 0$, then $y = |x| > 0$, so we can divide both sides by $y$:

$\dfrac{2y(y^2 + 1)}{y} = \dfrac{5y^2}{y}$

$2(y^2 + 1) = 5y$

$2y^2 + 2 = 5y$

$2y^2 - 5y + 2 = 0$

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Using the quadratic formula where $a = 2$, $b = -5$, and $c = 2$:

$y = \small\dfrac{5 \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}$

$y = \small\dfrac{5 \pm \sqrt{25 - 16}}{4} = \dfrac{5 \pm \sqrt{9}}{4} = \dfrac{5 \pm 3}{4}$

This gives us:

$y = \small\dfrac{5 + 3}{4} = \dfrac{8}{4} = 2$

$y = \small\dfrac{5 - 3}{4} = \dfrac{2}{4} = \dfrac{1}{2}$

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Now we convert back to $x$ values:

When $y = 2$:

Since $y = |x| = 2$, we have $x = 2$ or $x = -2$

Both are integers.

When $y = \dfrac{1}{2}$:

Since $y = |x| = \dfrac{1}{2}$, we have $x = \dfrac{1}{2}$ or $x = -\dfrac{1}{2}$

These are not integers, so we reject them.

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The integer solutions are: $x = 0, 2, -2$

Therefore, the number of integer solutions is 3.