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this should be simple. I am helping my son with one assignment but I simply cannot solve it. I really exhausted ideas. The problem is: prove that $\displaystyle {(m^2 + 5m)(m^2 + 5m + 10) + 24}$ can be divided by 24. Any hint?

user120250
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  • Well for zero it is one. The expression is equal to 1. – user120250 Jan 18 '16 at 17:45
  • I assume the question means "if $m$ is an integer, show that $(m^2+5m)(m^2+5m+10)$ is always divisible by $24$", no? – lulu Jan 18 '16 at 17:47
  • ror $m=0$ it is ok the statement – Dr. Sonnhard Graubner Jan 18 '16 at 17:47
  • Yes for m is integer. – user120250 Jan 18 '16 at 17:48
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    Hint: to check this, you just need to check that it is divisible by $3$ and $8$. Both of those are very quick checks. – lulu Jan 18 '16 at 17:49
  • Exactly, multiple of 24. – user120250 Jan 18 '16 at 17:49
  • OP: Kf-Sansoo gave an excellent answer. The idea is to factor your expression into something until it becomes clear that your expression is divisible by 24 regardless of the values you use for $m$. Thus, Kf-Sansoo rewrote your original expression as a difference of squares, $A^2-B^2$, and noted that $A^2-B^2=(A-B)(A+B)$. Also, both of these expressions then factored into the final expression he gave. Note that the final expression will always be divisible by 24 whenever you have an integer input $m$. If $m=-1,-2,-3,-4$, then the expression evaluates to $0$ which is, of course, divisible by 24. – Daniel W. Farlow Jan 18 '16 at 18:05

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$A = (m^2+5m + 5)^2 - 1 = (m^2 + 5m + 6)(m^2 + 5m + 4) = (m+1)(m+2)(m+3)(m+4)$ divisible by $4! = 24$.

DeepSea
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