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$a$ is correct as $n=2m+1 \text{(say)} \Rightarrow n^2=4m^2+4m+1\equiv1(\mod 8),\ m\ge1$

$b$ is true ingeneral, say $n=5, m=3$, for $c$ I dont know how to prove, please help

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If this is homework, tag it accordingly... –  draks ... Oct 4 '12 at 9:17
Exact duplicate of this question, since the question concerns only (c). See the answers there. –  Bill Dubuque Oct 4 '12 at 15:33

4 Answers 4

up vote 1 down vote accepted


(b)$(2x+1)^2+(2y+1)^2=4(x^2+y^2+x+y)+2\equiv2\pmod 4$

But $(2z)^2\equiv 0\pmod 4,(2z\pm1)^2\equiv1\pmod 4$

(c)$$\frac{n^5}5+\frac{n^3}5+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n 5+\frac n 3$$


Now using Femat's little theorem , prime $p\mid (n^p-n),$ so $5\mid (n^5-n)$ and $5\mid (n^3-3)$

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This answer will require a little work at the end. Since

$$\frac{n^5}{5}+ \frac{n^3}{3} + \frac{7n}{15}= \frac{1}{15}(3n^5 + 5n^3 +7n),$$

the expression above is an integer if and only if $3n^5 + 5n^3 +7n$ is divisible by $15$.

Then it is enough to check the finite number of choices: $n\in\{1,...,14\}$, i.e. a representative for every congruence class of numbers modulo 15. But we can restrict it even more. Indeed instead of checking $\{8,...,14\}$ we can check $\{-7,...,-1\}$ because they are congruent modulo 15. And we observe that $3(-n)^5 +5(-n)^3 +7(-n)=-(3n^5 + 5n^3 +7n)$.

So, at the end, we only have to check the following choices $n\in\{1,...,7\}$.

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Nice. I guess that checking the divisibility by 3 and by 5 separately could make it even easier (less cases, simpler expressions). –  maaartinus Oct 4 '12 at 13:03
Indeed, you are totally correct! It didn't come to my mind... –  Giovanni De Gaetano Oct 4 '12 at 13:35

By induction on $n$. For $n=1$ the result clearly holds. Suppose the result holds for $n$ and consider $\frac{(n+1)^5}{5}+\frac{(n+1)^3}{3}+\frac{7(n+1)}{15}=i+\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}$ where $i=1+2n+3n^2+3n^3+n^4$ is an integer. By the induction hypothesis we are done.

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Hint $\rm\displaystyle\ \frac{n^p}p + \frac{n^q}q + \frac{(pq\!-\!p\!-\!q)\,n}{pq}\ =\ \frac{n^p-n}p + \frac{n^q-n}q + n\,\in\,\Bbb Z\ $ for prime $\rm\,p,q\,$ by little Fermat.

Remark $\ $ That's essentially my answer to this duplicate question. See also the other answers there. More generally the same idea yields the following: $\,$ for $\rm\:a,b,c\in\Bbb Z,\:$ and primes $\rm\,p,q$

$$\rm\displaystyle \forall n\!:\ \frac{an^p}p + \frac{bn^q}q + \frac{cn}{pq}\ =\ \frac{a(n^p\!-\!n)}p + \frac{b(n^q\!-\!n)}q + \frac{(aq\!+\!bp\!+\!c)n}{pq}\,\in\,\Bbb Z \iff pq\ |\ aq\!+\!bp\!+\!c$$

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