Prove that, for $k \in {\mathbb Z^+}$ $$k(k+1)(k+2)(k+3)$$is not a perfect square.
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Tags: number theory, equation, Perfect Square
IsoLyS
30.01.2016 16:30
Note that: $k(k+1)(k+2)(k+3)=(k^2+3k+1)^2-1$
algebra_star1234
26.03.2017 17:09
Factor as $(k^3 + 3k + 1)^2-1$ which is only a perfect square when $k^3 + 3k+ 1 = 1 \implies k=0,-3$, but $k>0$ so we are done.
WizardMath
26.03.2017 18:09
Erdos proved a result with square replaced by an nth power and 3 replaced by a positive integer. For the above: Observe that the above is $(k^2+3k+1)^2-1$.
basemfouda2002
26.03.2017 18:32
algebra_star1234 wrote:
Factor as $(k^3 + 3k + 1)^2-1$ which is only a perfect square when $k^3 + 3k+ 1 = 1 \implies k=0,-3$, but $k>0$ so we are done.
Could you please explain how you approached the factorisation?
algebra_star1234
26.03.2017 18:35
@above: $$k(k+1)(k+2)(k+3)$$$$=k(k+3)(k+1)(k+2) = (k^2+3k)(k^2+3k+2)$$$$=(k^2+3k+1-1)(k^2+3k+1+1) = (k^2+3k+1)^2 - 1^2$$where the last statement is from difference of squares.