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I found that the concatenation in base 10 of $2^{541456}-1$ and $2^{541455}-1$, gives a probable prime.
I also found about number $541456$ that:

$$5cdot(5^2+4^2+1+4^2+5^2+6^2)=(5^3+4^3+1+4^3+5^3+6^3)$$

the equation $acdot(2cdot a^2+2cdot b^2+c^2+d^2)=(2cdot a^3+2cdot b^3+c^3+d^3)$ besides $a=5, b=4, c=6, d=1$ has other non trivial solutions with a,b,c,d positive integers?

Your equation $a*(2*a^2+2*b^2+c^2+d)=(2*a^3+2*b^3+c^3+d)$ is equivalent to
$$2ab^2+ac^2+ad=2b^3+c^3+d$$
I find the below solutions assuming all of the variables are in the range $1$ to $9$. I just did a search.

This section of this Answer is responsive to the version of the Question which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d^2) = (2 a^3 + 2 b^3 + c^3 + d^3)$.

Notice that $c$ and $d$ are interchangeable, so we may impose $c leq d$.

One way to look at this is $b,c,d geq 1$ and $a =frac{2 b^3 + c^3 + d^3}{2 b^2 + c^2 + d^2}$, when that is an integer. This leads to the question, does that ever happen? Yes. Notice if $b = c = d$, this is $a = (4b^3)/(4 b^2) = b$, so $a = b = c = d$ is a solution for any choice of $a$.

begin{align*}
&(a,b,c,d) & text{giving} \
&(4,4,4,4) & 384 \
&(5,5,5,5) & 750 \
end{align*}

Cylindrical algebraic decomposition with variable order $c, d, a, b$, gives five cylindrical components. Exhibiting an integer point in a component or showing that a component has no integer points is not easy.

• 
  $c geq 1$, $d geq 1$, $a = b = frac{c^3 + d^3}{c^2 + d^2}$, when this is an integer. The fraction is always an integer when $c = d$ and is sometimes an integer otherwise. Examples: $(1,1,1,1)$, $(9,9,5,10)$
  


• 
  $d geq c geq 1$, $a > frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the least real root of $2x^3 -2ax^2 -a(c^2 +d^2) +c^3 + d^3$. Examples: $(9,10,5,5)$, $(17,18,2,14)$, and $(251,255,51,251)$.


• 
  $d geq c geq 1$, $a$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when those roots are integers. (In progress: determining whether those roots are ever integers.)


• 
  $d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the middle root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(4,2,3,5)$ and $(5,2,2,6)$.


• 
  $d geq c geq 1$, $r$ is the least real root of $8x^3 + 27(c^2+d^2)x - 27(c^3+d^3)$, $r < a < frac{c^3 + d^3}{c^2 + d^2}$, and $b$ is the largest root of $2 x^3 - 2 a x^2 -a(c^2 +d^2)+c^3+d^3$, when this root is an integer. Examples: $(5,4,1,6)$ and $(17,15,5,20)$.

The below is responsive to the first version of the Question, which asked for positive integral solutions to $a (2 a^2 + 2 b^2 + c^2 + d) = (2 a^3 + 2 b^3 + c^3 + d)$.

Well, $a = b = c = d = 1$ is pretty obvious, giving $6$ on both sides.

One infinite family is $a=b=c = 1$, $d$ any positive integer, giving $d+5$. There don't seem to be more with $a = 1$, but I don't know how to prove that.

Here are eight more.
begin{align*}
&(a,b,c,d) & text{giving} \
&(2, 1, 7, 243) & 604 \
&(2, 1, 100, 979998) & 1980016 \
&(2, 2, 8, 384) & 928 \
&(2, 2, 68, 305184) & 619648 \
&(2, 3, 19, 6155) & 13084 \
&(2, 41, 1, 131117) & 268976 \
&(2, 41, 26, 147342) & 302776 \
&(2, 41, 27, 149343) & 306884
end{align*}

Cylindrical algebraic decomposition, using the variable ordering $b,c,a,d$ of your equation gives three solution families:

• 
  $a = b = c = 1$ and $d geq 1$, e.g., $(1,1,1,1)$ and $(1,1,1,108)$.


• 
  $b = 1$, $c geq 2$, $1 < a < frac{c^3+2}{c^2+2}$, $d = frac{c^3 - a c^2 - 2a + 2}{a-1}$, when $d$ is an integer, e.g., $(2,1,3,7)$ and $(109,1,126,2497)$.


• 
  $b geq 2$, $c geq 1$, $1 < a < frac{c^3+2b^3}{c^2+2b^2}$, $d = frac{c^4 - a c^2 + 2 b^3-2ab^2}{a-1}$, when $d$ is an integer, e.g., $(5,2,6,3)$ and $(3,3,100,485,000)$.

(The variable ordering $a,b,c,d$ gives more families, some of which appear to be empty, and most of which require bounds on $b$ and $c$ that are somewhat complicated algebraic functions of the prior variables. The previous version(s) of this answer used that ordering, but the new ordering is used to aid readability.)

An obvious solution lies where a,b,c,d all equal 1 as:
$$1cdot(2cdot1^2+2cdot1^2+1^2+1) = (2cdot1^3+2cdot1^3+1^3+1)$$