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Let $k > 3$ be an integer. Show that it is not possible for $k$ prime numbers, each greater than $k$, to be in an arithmetic progression with a common difference less than or equal to $k+1$.

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    First of all : let $p_{1}$ - first membe of ap. Then others write down as : $p_{1} + d, p_{1}+d + d , \dots , p_{1} + (k-1)d$, then $d$ couldn't be odd. – 2017-01-14
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    I know proof...using mod(d-1)...but,i want another one.... not using mod(d-1)....where,d is c.d. – 2017-01-14
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    Do you know about primitive residue classes modulo n? I recomend you to show that's this task is unsolvable for $d = k+1$ and then notice that $p_{1} , \dots , p_{1}+(k-1)d$ is system of residues modulo d, so there must be residue = $0$. – 2017-01-14

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Consider A.P. : $p_{1}, p_{1} + d, \dots , p_{1}+d(k-1)$. Now consider all $p_{i}\pmod{d-1}$.

Suppose $(p_{1},dm) = 1$, for $m = 1\dots k-1$. If that's not true then we have that some of $p_{i}$ isn't prime.

We think that all f $p_{i}$ are primes. Now this sequence is full system of residues modulo $d-1$. What does it mean ? It means that $p_{i} \pmod{d-1} = 0\dots d-2$ for all $i$. So there must be $j$:$p_{j} \equiv 0 \pmod{d-1}$, so $p_{j} = a(d-1)$, for $a\in\mathbb{N}$. Why this is true ? Suppose $p_{1}+dm \pmod{d-1} = p_{1} + dn\pmod{d-1}$, then $n\equiv m\pmod{d-1}$, so $m = n$. So there is exist $j$ : $p_{j}$ isn't prime.

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    Can you suggest some sources which contains this level of questions??and do tou know source of this question?thank u – 2017-01-16
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$\boxed{\text{Primes in AP}.}$

Lemma: About complete set of residue.

If $gcd(n,m)=1$ and $t_{i}=a+(i)m$ then, $t_1,t_2,...,t_{n}$ (mod$n$) is permutation of $0,1,2,...,(n-1)$.

[Proof:] If not,then by PHP, $t_i$ and $t_j$ will have same remainder when divided by $n$, i.e., $t_i \equiv t_j$ (mod$n$). So $n$ divides $t_i-t_j=(m(i-j))$ but gcd=1, so $n$ divides $(i-j)$, but $i-j$ is less than $n$ Contradiction.

Let $k$ primes in AP, and $(d-1) \leq k$ and $p>k$

$p,p+d,....,p(k-1)d$

Since, $gcd(d,d-1)=1$ (In this case $n=d-1$ and $m=d$)

Consider, $p,p+d,....,p+(k-1)d$ mod($d-1$). This series contains $0,1,2,...,(d-2)$ and some more.(since $(d-1) \leq k$, i.e., we have ($k$) more terms than required for complete set of $0,1,2,...,(n-1)$,here $n=d-1$,)

So there exists one $p_i$ such that, $(d-1)$ divides $p_i$. ,,, but $p_i$ is prime then we have, $p_i=d-1$ but $d-1$ is less than or equal to $k$ and $p_i$ is strictly graeter than $k$. Contradiction.