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I am reading the book 'Essential Engineering Mathematics' by Alan Jeffrey. There is one example that asks to prove, by mathematical induction, that $\forall n\in \mathbb{N}$ the number $$ 2^{n+2}\times 3^n+5n-4 $$ is divisible by 25. I do not have any issues with the proof (or similar proofs). My question is may be a dubious one: how can one arrive at asserting a proposition like the above in the first place?

Dimitris
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    It is easy to see the number is divisible by $5$: $2^{n+2}\times 3^n = 4\times 6^n$; since $6$ leaves a remainder of $1$ when dividided by $5$, this number leaves a remainder of $4$ when divided by $5$, so $2^{n+2}\times 3^n -4$ is a multiple of $5$, and hence so is $2^{n+2}\times 3^n + 5n - 4$. Modulo $25$ you have to work a bit more more; the powers of $6$ modulo $25$ cycle as $6\mapsto 11\mapsto 16\mapsto 21\mapsto 1\mapsto 6$, so you can consider the different possible values of $n$ in turn and verifying the result is divisible by $25$. – Arturo Magidin Dec 15 '22 at 22:51
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    You can also note that $(1+5)^n \equiv 1 + 5 n \mod 5^2$ by the binomial theorem. – Robert Israel Dec 15 '22 at 22:56
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    You mean: How in the world could anybody ever figure out that $2^n+7^n+8^n+18^n+19^n+24^n=3^n+4^n+12^n+14^n+22^n+23^n \text{ for } n =0,1,...,5$ Some things remain a mystery to me. – Marius S.L. Dec 15 '22 at 23:15
  • Such divisibilities (or congrunces) arise immediately from the Binomial Theorem (or from recurrences or closely related results) - see here and its links in the linked dupe for a general form of such. Write it as $,f_n = 4(6^n)+5n-5,$ then apply the linked Theorem. – Bill Dubuque Dec 16 '22 at 03:30

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