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Pick out the true statement(s):
(a) If $n$ is an odd positive integer, then $8$ divides $n^2 – 1$.
(b) If $n$ and $m$ are odd positive integers, then $n^2+m^2$ is not a perfect square.
(c) For every positive integer $n$, $n^5/5+n^3/3+7n/15$ is an integer.

(a) is true by induction. But I can not verify the others. Please help me somebody

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(a) can also be proved without induction.

$$(2c+1)^2=8\frac{c(c+1)}2+1\equiv1\pmod 8$$

(b)$$(2a+1)^2+(2b+1)^2=4\{a^2+a+b^2+b\}+2\equiv2\pmod 4$$

But any number $m\equiv0,1,2,3\pmod 4\implies m^2\equiv0,1\pmod 4$

(c) Using Fermat's Little Theorem, prime $p\mid(n^p-n) $ for all integer $n$.

So, $$5\mid (n^5-n) \text{ and } 3\mid(n^3-n)$$

Hence, $$\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\left(\frac n3+\frac n5+\frac{7n}{15}\right)=\frac{n^5-n}5+\frac{n^3-n}3+n$$

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You can also make use of (a) in (b): From (a) you know that $n^2+m^2-2$ is a multiple of $8$, hence $n^2+m^2$ is a multiple of $2$, but not of $4$. If $n^2+m^2$ were the square of an even number, it would be a multiple of $4$ - which it isn't. If $n^2+m^2$ were the square of an odd number, it would be odd (again by (a)) - which it isn't. – Hagen von Eitzen Jan 16 '13 at 16:20
(+1) For the c) part – Inceptio Apr 12 '13 at 9:18

(b) $(2a+1)^2+(2b+1)^2=4(a^2+a+b^2+b)+2\equiv 2\pmod{4}$

But any number $m≡0,1,2,3\pmod 4 \implies m^2\equiv 0,1\pmod 4$.

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See the editing help on how to format your answers. – Mårten W Apr 12 '13 at 9:34

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