A question about the number 541456

Question

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:

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

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

Answer 1

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

Answer 2

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.

Answer 3

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. Having exhausted my tools for determining whether this components has integral points, I am unable to find any or prove that there are none. (Taking intersections with CADs having different variable orders splits this component into pieces, for instance $a \geq 2$, $b = 2 a/3$, $c = 2 a/3$, and $d$ is the least real root of $9 x^3 - 9 a x^2 - 4 a^3 = 0$. It's unclear whether that root is ever an integer for an integer $a$ divisible by $3$.
• $d \geq c \geq 1$, $r$ 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.)