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Fermat's Two-Square Theorem:
Given a prime $ p$, there exist integers $ a, b$ such that $ a^2 + b^2 = p$ iff $ p = 2$ or $ p \equiv 1 \bmod 4$. Consequently, a number $ n$ is expressible in the form $ a^2 + b^2$ iff the primes congruent to $ 3 \bmod 4$ in its prime factorization each divide $ n$ an even number of times.

But for example, if we take $49$ whose prime factorisation is $7^2$, all the primes congruent to $3 \bmod 4$ have their power as even, so $49$ should be expressible as sum of squares of two integers, although $49$ can't be expressed as $a^2 + b^2$.

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    $\begingroup$ Note that $49=0^2+7^2$ $\endgroup$ – Mark Bennet Oct 3 '15 at 11:30
  • $\begingroup$ that was helpful . $\endgroup$ – rahul_mishra01 Oct 3 '15 at 17:49
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As Mark Bennet points out, $49$ can be written as the sum of two squares, namely $49 = 7^2+0^2$.

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Since $49$ is obviously equal to the sum of two squares, $0^2 + 7^2$, perhaps Rahul meant to say that there exist no integers for $a$ and $b$ in the equation $a^2 + b^2 = 49$ such that $(a,b) > 0$. Perhaps it would be better to say that there exists no such "natural number" for $a$ and $b$ since $0$ is not a natural number. In fact, all numbers equal to and below $0$ are not natural numbers, but we know for any number $n$, $n^2 = (-n)^2$.

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