For any integer n, $78n^2+72n+2018$ is expressed as sum of four perfect cubes.

Question

For any integer n, $$78n^2+72n+2018$$ is expressed as sum of four perfect cubes.

Is this possible that any integer can fit into this expression with 4 perfect cubes?

Answer 1

Try this expression: $$78n^2+72n+2018 = (n+a)^3+(n+b)^3+(-n+c)^3+(-n+d)^3.$$ Then $$\left\{\begin{array}{l} 78n^2=3n^2(a+b+c+d);\\ 72n = 3n(a^2+b^2-c^2-d^2);\\ 2018 = a^3+b^3+c^3+d^3.\end{array}\right. $$ $$\left\{\begin{array}{l} a+b+c+d=26;\\a^2+b^2-c^2-d^2=24;\\ a^3+b^3+c^3+d^3=2018.\end{array}\right. $$

This system has integer solution $a=1, b=11, c=7, d=7$.

Therefore $$ 78n^2+72n+2018=(1+n)^3+(11+n)^3+(7-n)^3+(7-n)^3. $$

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Many numbers of the form $$78n^2+72n+2018$$ can be expressed as sum of up to four positive cubes too, but not all ($n=16$: $23138$, $n=22$: $41354$, $\ldots$).

Few examples:
$7$: $6344=2^3+ 8^3+ 12^3+ 16^3\quad (=8^3+18^3)$;
$8$: $7586=6^3+ 9^3+ 12^3+ 17^3$;
$9$: $8984=5^3+ 10^3+ 10^3+ 19^3$;
$10$: $10538=5^3+ 6^3+ 13^3+ 20^3\quad (=11^3+15^3+18^3)$;
$11$: $12248=3^3+ 3^3+ 3^3+ 23^3$;
$12$: $14114=3^3+ 4^3+ 15^3+ 22^3$;
$13$: $16136=2^3+ 2^3+ 19^3+ 21^3\quad (=14^3+14^3+22^3)$ ;
$14$: $18314=1^3+ 2^3+ 9^3+ 26^3$;
$15$: $20648=10^3+ 10^3+ 20^3+ 22^3$;

$17$: $25784\;(=4^3+ 11^3+ 29^3)$;
$19$: $31544=8^3+ 8^3+ 9^3+ 31^3$;
$20$: $34658=1^3+ 14^3+ 17^3+ 30^3$;
$21$: $37928=2^3+ 7^3+ 25^3+ 28^3$.