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I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$ We also have: $$13^2+43^2=2018$$

And we have:

$$2018=44^2+9^2+1^2$$

I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.

So we have a system of three Diophantine equations:

$$n=a^2+b^2$$

and $$n=c^2+d^2+e^2$$

and $$n=f^4+g^4+h^4+i^4$$

where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.

Is there a finite number of these numbers?

Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.

And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.

  • I think you first need to solve the system. $$a^2+b^2=c^2+d^2+e^2=f^2+g^2+h^ 2+i^2$$ Then find out when decisions $f,g,h,i - $ are the squares. You will need to solve another system of equations. – individ Dec 31 '17 at 15:23

3 Answers3

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There are an infinite number of such integers.

One way to show this is to start from the fact that there are an infinite number of Pythagorean triples $a^2 + b^2 = c^2$. Given any such triple, let $N$ be given by:

$$N = (ab)^4 + (bc)^4 + (ac)^4 + c^4$$

By an identity of Fauquembergue (1) we have (given $a^2 + b^2 = c^2$):

$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2$$

Hence:

$$N = (a^4 + a^2b^2 + b^4)^2 + (c^2)^2$$

Furthermore:

$$(c^2)^2 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 = (a^2 - b^2)^2 + (2ab)^2$$

Hence:

$$N = (a^4 + a^2b^2 + b^4)^2 + (a^2 - b^2)^2 + (2ab)^2$$

Reference:

1) Dickson L E History of the Theory of Numbers Vol 2 Ch XXII p 658

Adam Bailey
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    More generally, if $\color{blue}{x^4+y^4+z^4=t^2}$ then, $$x^4+y^4+z^4+c^4=t^2+(c^2)^2 = t^2+(a^2-b^2)^2+(2ab)^2$$ where $a^2+b^2=c^2$. The crucial detail is then the blue equation. I did a computer search and some small solutions were given by the Fauquembergue parametrization $F$. Some non-$F$ ones are, $$60^4 + 135^4 + 148^4 = 28721^2 \ 72^4 + 175^4 + 240^4 = 65441^2 \ 155^4 + 260^4 + 296^4 = 113241^2$$ It would be nice to know if there are other nice paramaterizations to the blue equation. – Tito Piezas III Jan 01 '18 at 06:36
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One infinite set is $2018k^4$ for any natural $k$. I strongly suspect that there are plenty more. Numbers that are a sum of three squares are very common. Numbers that are a sum of two squares are not so rare, so I would just start picking sums of four cubes and try to satisfy the other two. Another example is $$1^4+2^4+3^4+6^4=1394=2\cdot 17 \cdot 41 \equiv 4\pmod 8$$ so is a sum of two and three squares.

Ross Millikan
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  • My example shows it is not as $1394$ is smaller. I tried $1,2,3,4$ and $1,2,3,5$ without success. The smallest is $1^4+2^4+4^4+5^4=898$ – Ross Millikan Dec 31 '17 at 15:57
  • I deleted a comment when I saw that 1394 is also an example. –  Dec 31 '17 at 15:58
  • So you believe that there is an infinite number of families of these numbers? –  Dec 31 '17 at 15:59
  • Yes, I believe so. I don't have a proof, but two of the four sets I tried provided solutions. – Ross Millikan Dec 31 '17 at 15:59
  • 2018 is also a sum of five squares, I discovered that a couple of minutes ago. –  Dec 31 '17 at 16:00
  • Every number is a sum of four squares, so being a sum of five is not special at all – Ross Millikan Dec 31 '17 at 16:14
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    @RossMillikan: The smallest is $34$, since $$1^4+1^4+2^4+2^4 = 3^2+3^2+4^2 = 3^2+5^2 =34$$ He didn't specify that terms be distinct. In fact, the sequence starts as, $$34, 164, 194, 244, 274, 289, 369, 514, 544, 628, 673, 674, 738, 769, 788, 898,\dots$$ – Tito Piezas III Dec 31 '17 at 16:45
  • @TitoPiezasIII: You are right, I was assuming the terms had to be distinct. – Ross Millikan Dec 31 '17 at 16:46
  • @TitoPiezasIII Can´t take my eyes off successive 673 and 674 –  Dec 31 '17 at 23:20
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Sum of 4 distinct fourth powers: $$2018 =2^4 + 3^4 + 5^4 + 6^4$$

as well as sum of 12 consecutive squares: $$2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2$$

and "smallest number equal to the product of two primes which is also equal to the sum of 33 distinct primes": http://oeis.org/A102238

and the third least k > 0 such that the nextprime(k$\times$primorial(n)) - k$\times$primorial(n) is composite: http://oeis.org/A071771

Tito Piezas III
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