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Every four consecutive integers contains one which cannot be written as sum of two squares.

Could anyone advise me how to prove the statement? Do I use Jacobi's two squares theorem? Hints will suffice, thank you.

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    $\begingroup$ Out of four consecutive integers, one is congruent to $0$ modulo 4, one is congruent to $1$ modulo 4, one is congruent to $2$ modulo 4, and one is congruent to $3$ modulo 4. $\endgroup$ – Lord Shark the Unknown Oct 30 '18 at 18:26
  • $\begingroup$ Hint: What are the squares mod 4? What are the possible values for sums of two squares mod 4? $\endgroup$ – JavaMan Oct 30 '18 at 18:27
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Hint:

  • There is a number of the form of $4k+3$.

  • You might like to explore $x^2 \pmod{4}$.

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You may consider the following as a hint:

Squares are of the form $4k$ or $4k+1$ "easy to verify $(2k)^2$,$(2k+1)^2$". Now, what are the possible forms of the summation of these numbers?

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    $\begingroup$ It seems that numbers of the form $4k+3$, which occurs in every four consecutive integers, cannot be written as sum of two squares. $\endgroup$ – Alexy Vincenzo Oct 31 '18 at 1:03
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    $\begingroup$ which proves the statement. is not it? $\endgroup$ – Maged Saeed Oct 31 '18 at 3:04
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One further hint: Consider that Pythagorean triples $x^2+y^2=z^2$ can be found as $x=m^2-n^2,\space y=2mn,\space z=m^2+n^2$.

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