Use theory of congruence to prove.....

Question

Use the theory of congruence to prove that $17|(2^{3n+1} +3\times5^{2n+1})$ for all integer $n\geq1$ $(2^{3n+1} +3\times5^{2n+1})$=$2\times8^n+15\times25^n$ =$17\times8^{n-1}+374\times25^{n-1}+25^{n-1}-8^{n-1}$

=$25^{n-1}-8^{n-1}$ =$8^{n-1}-8^{n-1}$ [since $25\equiv 8\mod 17)$ =$0\mod 17$

$0$ is divisible by $17$

Is this correct?

@HagenvonEitzen OP probably meant $3\cdot 5$, and an edit is waiting to be approved. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jun 21 '18 at 06:27
Note that $25^{n-1} - 8^{n-1}$ is always a multiple of $25-8 = 17$ ($a-b$ always divides $a^n - b^n$). — Sarvesh Ravichandran Iyer, Jun 21 '18 at 06:32

score 0 · Answer 1 · answered Jun 21 '18 at 06:56

0

Starting from here: $$\begin{align}2\times8^n+15\times25^n&\equiv2\times8^n+15\times (17+8)^n\equiv\\ &\equiv2\times 8^n+15\times 8^n\equiv\\ &\equiv17\times 8^n\equiv 0 \pmod{17}.\end{align}$$

answered Jun 21 '18 at 06:56

farruhota

31,482

Use theory of congruence to prove.....

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