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Prove that there are arbitrarily long sequences of consecutive integers, none of which can be written as the sum of two perfect squares.

First few numbers are $3,6,7,11,12,14,15,19,21,22,23,24,27,28,30,31,33,35,38,39, \cdots$

Sums of squares can only be of the form $4k$, $4k+1$ and $4k+2$. So can we use this idea to prove the proposition?

I didn't find a logical sequence. Can anyone provide some hints to proceed with this?

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A number is a sum of two squares if and only if all primes $3$ mod $4$ appear only to even powers in its prime factorisation.

So we want arbitrarily long sequences of consecutive integers such that for each $n$ in the sequence, some prime $p \equiv 3 \pmod{4}$ appears to an odd power in the prime factorisation of $n$.

There are infinitely many primes $3$ mod $4$; say the $n$th is $p_n$.

Then we can guarantee that a number is divisible by $p_n$ but not $p_n^2$ by insisting that it be of the form $m$ where $m \equiv p_n \pmod{p_n^2}$.

By the Chinese remainder theorem, we can find $a$ such that $a+1 \equiv p_1 \pmod{p_1^2}$, $a+2 \equiv p_2 \pmod{p_2^2}$, and so on up to $a+k \equiv p_k \pmod{p_k^2}$. This is a sequence of $k$ consecutive integers, none of which is a sum of two squares.

For example, if $k=5$, we wish to find $a$ such that $$a+1 \equiv 3 \pmod{9} \\ a+2 \equiv 7 \pmod{49} \\ a+3 \equiv 11 \pmod{121} \\ a+4 \equiv 19 \pmod{361} \\ a+5 \equiv 23 \pmod{529}$$

The smallest such $a$ is $1789983137$ (and it is unique with this property modulo $10190296809$). Therefore $$ 1789983138,1789983139,1789983140,1789983141, 1789983142$$ is such a sequence. I emphasise that this is only one working sequence; in fact the smallest is $75, 76, 77, 78, 79$.

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  • $\begingroup$ Could you please include an example for $k=5$? $\endgroup$ – TheRandomGuy Mar 25 '16 at 13:42
  • $\begingroup$ @Dhruv Done. I doubt it'll be very helpful though :P $\endgroup$ – Patrick Stevens Mar 25 '16 at 13:50
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    $\begingroup$ You don't really need $a+4$ congruent with 19 mod 361. It's already congruent with 6 mod 9. Drop that condition, replace 23 with 19 (and swap the squares of those two numbers) for $a+5$, and see how much smaller the numbers get. $\endgroup$ – Oscar Lanzi Mar 31 '16 at 0:02
  • $\begingroup$ @OscarLanzi That yields an example with $a=24907978$. Well spotted. $\endgroup$ – Patrick Stevens Mar 31 '16 at 7:12

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