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I'll state the question here:

If "$P(n): 49^n + 16^n + k$ is divisible by $64$ for all $n \in N$" is true, then what is the least negative integral value of $k$?

This is how I tried to solve it:

$P(1): 49^1 + 16^1 + k$ is divisible by $64$ for all $n \in N$

i.e., $65 + k = 64p$, where $p \in Z$

i.e., $k = 64p - 65$

i.e., $k = -1$, for $p = 1$ since $k$ is the least negative value

This is what the answer according to my book is and this is how it was solved in the book. But how do we know that this is the correct answer? The smallest negative value of $k$ is $-1$ when I consider $P(1)$. How can I say that $k = -1$ for all natural numbers?

Is this the correct solution to the problem? Or should I replace $k$ by $-1$ in $P(n)$ and then prove it by induction (that is what the chapter in my book is about)? If I am able to prove it (that I will be since this is actually the answer) by induction, then this is the answer. Otherwise, there's some other answer.

Any help would be appreciated.

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    Perhaps use $ p = 1 mod (4) $ identity on the graph $ k = 64p - 65 $ to show that it is divisible at multiples of four along the gradient of 64. – 2017-02-09
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    $49^2+16^2-1$ is not divisible by $64$. – 2017-02-09
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    There is no smallest k for any individual $n$. It's just that the only $k$ that is in common with all $n$ is $-1$. – 2017-02-09
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    @kingW3 You are right. Then what is the correct answer? – 2017-02-16
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    No such $k$ exists. – 2017-02-16
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    @kingW3 Can you please elaborate why? And please don't use modular arithmetic like the answer below because I am not familiar with it. – 2017-02-21

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No such $k$ exists.

Notice that we would need $k\equiv -(49^n+16^n)\bmod 64$ for all $n\in\mathbb N$.

However $49^n+16^n$ already takes two different values $\bmod 64$ for $n=1,2$.

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    Sorry. I haven't studied modular arithmetic. Can you please explain it in some other way? – 2017-02-11