Prime and Factorization

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dshasan
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Prime and Factorization

Unread post by dshasan » Tue Jan 17, 2017 9:22 pm

Find all prime numbers $p$ for which $5p+1$ is a perfect square.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

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Thanic Nur Samin
Posts:176
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Re: Prime and Factorization

Unread post by Thanic Nur Samin » Tue Jan 17, 2017 10:27 pm

$$5p+1=x^2$$
$$\Rightarrow 5p=(x+1)(x-1)$$

Case 1: $x-1=1, x+1=5p$

This implies $5p=3$, impossible.

Case 2: $x-1=5,x+1=p$

This gives us solutions $(x,p)=(6,7)$

Case 3: $x-1=p,x+1=5$

This gives us solutions $(x,p)=(4,3)$

So the only answers are $p=3,7$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

dshasan
Posts:66
Joined:Fri Aug 14, 2015 6:32 pm
Location:Dhaka,Bangladesh

Re: Prime and Factorization

Unread post by dshasan » Tue Jan 17, 2017 10:33 pm

Thanic Nur Samin wrote:
$$5p+1=x^2$$
$$\Rightarrow 5p=(x+1)(x-1)$$

Case 1: $x-1=1, x+1=5p$

This implies $5p=3$, impossible.

Case 2: $x-1=5,x+1=p$

This gives us solutions $(x,p)=(6,7)$

Case 3: $x-1=p,x+1=5$

This gives us solutions $(x,p)=(4,3)$

So the only answers are $p=3,7$.
Same as my solution :D
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

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Thanic Nur Samin
Posts:176
Joined:Sun Dec 01, 2013 11:02 am

Re: Prime and Factorization

Unread post by Thanic Nur Samin » Tue Jan 17, 2017 10:54 pm

Another solution:
$$5p+1=x^2$$
$$\Rightarrow x^2\equiv 1(\operatorname{mod} 5)$$
$$\Rightarrow x\equiv \pm 1$$
$$\Rightarrow x=5y\pm 1$$

So, $5p+1=(5y\pm 1)^2$
$$\Rightarrow 5p+1=25y^2\pm 10y+1$$
$$\Rightarrow p=5y^2\pm 2y$$
$$\Rightarrow p=y(5y\pm 2)$$

Since $5y\pm 2=\pm 1$ is nonsense, we get that $y=\pm 1$ and thus $p=7,3$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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