## N = A^3 + B^3 + C^3 + D^3 = E^3 + F^3 + G^3

To find integers   $N$   such that

$N = A^3 + B^3 + C^3 + D^3 = E^3 + F^3 + G^3$

with one more term on the left than on the right

Note the identity:

$(3 a + 3 b)^3 + (2 a + 4 b)^3 + a^3 + b^3$
$= 36 a^3 + 129 a^2 b + 177 a b^2 + 92 b^3$
$= a^3 (27 + 8 + 1) + a^2 b (108 + 12 + 9) + a b^2 (144 + 6 + 27) + b^3 (64 + 1 + 27)$
$= 27 a^3 + 108 a^2 b + 144 a b^2 + 64 b^3$
$+ 8 a^3 + 12 a^2 b + 6 a b^2 + b^3$
$+ a^3 + 9 a^2 b + 27 a b^2 + 27 b^3$

$= (3 a + 4 b)^3 + (2 a + b)^3 + (a + 3 b)^3$

we have

$(3 a + 3 b)^3 + (2 a + 4 b)^3 + a^3 + b^3 = (3 a + 4 b)^3 + (2 a + b)^3 + (a + 3 b)^3$

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## (1,2,3,…,27) separated into 3 sets

$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27$

separated into 3 sets:
$1, 6, 8, 12, 14, 16, 20, 22, 27$
$2, 4, 9, 10, 15, 17, 21, 23, 25$
$3, 5, 7, 11, 13, 18, 19, 24, 26$

the sum in any set is the same
$1 + 6 + 8 + 12 + 14 + 16 + 20 + 22 + 27 = 126$
$2 + 4 + 9 + 10 + 15 + 17 + 21 + 23 + 25 = 126$
$3 + 5 + 7 + 11 + 13 + 18 + 19 + 24 + 26 = 126$

sum of squares of the numbers is the same
$1^2 + 6^2 + 8^2 + 12^2 + 14^2 + 16^2 + 20^2 + 22^2 + 27^2 = 2310$
$2^2 + 4^2 + 9^2 + 10^2 + 15^2 + 17^2 + 21^2 + 23^2 + 25^2 = 2310$
$3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 18^2 + 19^2 + 24^2 + 26^2 = 2310$

Take the numbers from   1 to 64,   separate them into 4 sets such that the sum and sum of squares (or cubes) of the numbers in any set are the same as for the other sets.

N.B.   The generalization is a bit difficult.

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## product of tangents

$\tan (\pi/3) = \sqrt{3}$

$\tan (\pi/5) \tan (2 \pi/5) = \sqrt{5}$

$\tan (\pi/7) \tan (2 \pi/7) \tan (3 \pi/7) = \sqrt{7}$

$\tan (\pi/9) \tan (2 \pi/9) \tan (3 \pi/9) \tan (4 \pi/9) = 3$

$\tan (\pi/11) \tan (2 \pi/11) \tan (3 \pi/11) \tan (4 \pi/11) \tan (5 \pi/11) = \sqrt{11}$

$\tan (\pi/13) \tan (2 \pi/13) \tan (3 \pi/13) \tan (4 \pi/13) \tan (5 \pi/13) \tan (6 \pi/13) = \sqrt{13}$

$\tan (\pi/15) \tan (2 \pi/15) \tan (3 \pi/15) \tan (4 \pi/15) \tan (5 \pi/15) \tan (6 \pi/11) \tan (7 \pi/15)$
$= \sqrt{15}$

the pattern continues.

Can you explain why?

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## x^n (y-z) + y^n (z-x) + z^n (x – y); n=1,2,3,4,5

$x (y-z) + y (z-x) + z (x-y) = 0$

$x^2 (y-z) + y^2 (z-x) + z^2 (x - y)$
$= x^2 y - x^2 z - x y^2 + x z^2 + y^2 z - y z^2$
$= (x-y) (x-z) (y-z)$

$x^3(y - z) + y^3(z - x) + z^3(x - y)$
$= x^3 y - x^3 z - x y^3 + x z^3 + y^3 z - y z^3$
$= (x-y) (x-z) (y-z) (x+y+z)$

$x^4 (y - z) + y^4 (z - x) + z^4 (x - y)$
$= x^4 y - x^4 z - x y^4 + x z^4 + y^4 z - y z^4$
$= (x-y) (x-z) (y-z) (x^2 + x y + x z+ y^2 + y z + z^2)$

So

$x^2 (y-z) + y^2 (z-x) + z^2 (x - y) = (x-y) (x-z) (y-z)$
$x^3(y - z) + y^3(z - x) + z^3(x - y) = (x-y) (x-z) (y-z) (x+y+z)$
$x^4 (y - z) + y^4 (z - x) + z^4 (x - y) = (x-y) (x-z) (y-z) (x^2 + x y + x z + y^2 + y z + z^2)$

Factorize    $x^5 (y-z) + y^5 (z-x) + z^5 (x - y)$

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## N = 1 + x + … + x^m = 1 + y + … + y^n

Find integers   $N$   that have 2 representations as a sum of powers

$N = 1 + x + ... + x^m = 1 + y + ... + y^n$

where   $x$   and   $y$   are positive integers.

$x \ge 2$    and    $y \ge 2$

E.g.

$31 = 1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 5 + 5^2$

$8191 = 1 + 2 + 2^2 + ... + 2^{12} = 1 + 90 + 90^2$

Can you find more examples?

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## N = A^2 + B^2 + C^2 = X^2 + Y^2 + Z^2

Part #1

To find integers   $N$   such that

$N = A^2 + B^2 + C^2 = X^2 + Y^2 + Z^2$

Let’s take six integers
$x+y, b, c$    and    $x, y, b+c$

$(x+y)^2 + b^2 + c^2 = b^2 + c^2 + x^2 + 2 x y + y^2$

$x^2 + y^2 + (b+c)^2 = b^2 + 2 b c + c^2 + x^2 + y^2$

$(x+y)^2 + b^2 + c^2 = x^2 + y^2 + (b+c)^2$    if   $x y = b c$

E.g.

x + y = 2 + 3,    b = 1,    c = 6,    x = 2,    y = 3,    b + c = 7

and    $x y = 2*3 = 1*6 = b c$

$62 = 5^2 + 1^2 + 6^2 = 2^2 + 3^2 + 7^2$

——————————————

Part #2

From an integer N with 2 representations as a sum of 3 squares to an integer M with 2 representations as a sum of 4 squares

If    $N = a^2 + b^2 + c^2 = x^2 + y^2 + z^2$

then,   consider the two sets of integers:
$a, b, c + k z, kc$    and    $x, y, z + kc, k z$

$a^2 + b^2 + (c + kz)^2 + (kc)^2 = a^2 + b^2 + c^2 k^2 + c^2 + 2 c k z + k^2 z^2$

$x^2 + y^2 + (z + kc)^2 + (kz)^2 = x^2 + y^2 + c^2 k^2 + 2 c k z + k^2 z^2 + z^2$

It’s clear that

$M = a^2 + b^2 + (c + kz)^2 + (kc)^2 = x^2 + y^2 + (z + kc)^2 + (kz)^2$

——————————————

Part #3

$(ab + cd)^2 + (bc)^2 + (ad)^2$
$= a^2 b^2 + a^2 d^2 + 2 a b c d + b^2 c^2 + c^2 d^2$
$= (b^2 c^2 + a^2 d^2 + 2 a b c d) + a^2 b^2 + c^2 d^2$
$= (b c + a d)^2 + (a b)^2 + (c d)^2$

So

$(a b + c d)^2 + (b c)^2 + (a d)^2 = (b c + a d)^2 + (a b)^2 + (c d)^2$

If we have consecutive integers:

$a = n - 1$ ,   $b = n$ ,   $c = n + 1$ ,   $d = n + 2$

$((n-1)n + (n+1)(n+2))^2 + (n(n+1))^2 + ((n-1)(n+2))^2$
$= 6 n^4+12 n^3+10 n^2+4 n+8$

And,

If $a = n - 2$ ,   $b = n$ ,   $c = n + 2$ ,   $d = n + 4$

$((n - 2)n + (n + 2)(n + 4))^2 + (n(n + 2))^2 + ((n - 2)(n + 4))^2$
$= 2 (3 n^4 + 12 n^3 + 20 n^2 + 16 n + 64)$

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## N = A^2 + B^2 + C^2 + D^2 = E^2 + F^2 + G^2

To find integers   $N$   such that

$N = A^2 + B^2 + C^2 + D^2 = E^2 + F^2 + G^2$

with one more term on the left than on the right

Note the identity:

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

we have

$(a+b+c) + (a+b-c) + (a-b+c) + (-a+b+c) = 2 a + 2 b + 2 c$

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

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