Not enough pairs

Number Theory Level 3

$\large n!+1=m^2$

Let the pairs $(m,n)$ be positive integers such that the equation above is fulfilled.

How many pairs of $(x,y)$ there are such that $m<100$ ?

The answer is 3.

1 solution

Abdeslem Smahi
Jul 14, 2015

Actually there is only three pairs known till know have this property even for $m>100$ these pairs are called brown numbers and they are $(5,4);(11,5);(71,7)$ there is no such a proof that there is only three pairs but the mathematicians conjectured that those are the only ones.

Moderator note:

Your solution did not show any relevant working. What are the potential cases we should check that satisfy all these condition?

Hint : Focus on the factorial sign.

This is known as the Brocard's Problem . And it is still an open question.

actually I used This:

because $m<100$ so $m²<10000$ so $n!+1<10000$

$\implies n<8$

for $n=0 \implies n!+1=2$ not a perfect square

for $n=1 \implies n!+1=2$ not a perfect square

for $n=2 \implies n!+1=3$ not a perfect square

for $n=3 \implies n!+1=7$ not a perfect square

for $n=4 \implies n!+1=25=5^2$ so we have the pair $(5,4)$

for $n=5 \implies n!+1=121=11^2$ another pair $(11,5)$

for $n=6 \implies n!+1=721$ not a perfect square

for $n=7 \implies n!+1=5041=71^2$ the last pair $(71,7)$

so we have $3$ pairs $(5,4);(11,5);(71,7)$

Abdeslem Smahi - 5 years, 11 months ago

Not enough pairs

The answer is 3.

1 solution

Moderator note:

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