Find a pair of positive integers to get a square number !! I have tried to solve this for 2 hours

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nakamoto2305

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Find the pair of positive integers [MATH]x,y[/MATH] such that [MATH]k[/MATH] is the smallest possible and [MATH](n^2+1)((n+k)^2+1)[/MATH] is a square number.
I can easily prove that [MATH]k[/MATH] must be an even number, and I also found out the answer is [MATH](n;k)=(1;6)[/MATH].
I can prove that with k=2, there is no solution, but I still confuse with k=4; btw, this way is not good because it looks like I guess the answer and just prove there is no solution for n with all value [MATH]k[/MATH] smaller than [MATH]6[/MATH] .
So I try another way that can prove clearly k=6 is the smallest that can give a solution for this problem, can u help me?
And sorry for my bad grammar.
 
nakamoto2305 said:
Find the pair of positive integers [MATH]x,y[/MATH] such that [MATH]k[/MATH] is the smallest possible and [MATH](n^2+1)((n+k)^2+1)[/MATH] is a square number.
I can easily prove that [MATH]k[/MATH] must be an even number, and I also found out the answer is [MATH](n;k)=(1;6)[/MATH].
I can prove that with k=2, there is no solution, but I still confuse with k=4; btw, this way is not good because it looks like I guess the answer and just prove there is no solution for n with all value [MATH]k[/MATH] smaller than [MATH]6[/MATH] .
So I try another way that can prove clearly k=6 is the smallest that can give a solution for this problem, can u help me?
And sorry for my bad grammar.
Click to expand...
You say for k = 2 , there is no solution.

Please share your work.

What "class" has assigned this problem?
 
Subhotosh Khan said:
You say for k = 2 , there is no solution.

Please share your work.

What "class" has assigned this problem?
Click to expand...

with k=2; we prove that [MATH](n^2+1,(n+2)^2+1)=1[/MATH] or [MATH]2[/MATH], now we suppose that p is one prime common divisor of they and [MATH]p>2[/MATH] => [MATH]p | (n^2+1) [/MATH] and [MATH]p | (n+2)^2+1) = n^2+1+4(n+1) => p | 4(n+1) [/MATH] , if p>2 so [MATH]p | (n+1) [/MATH] => \[n \equiv - 1(\bmod p) \Rightarrow {n^2} \equiv 1(\bmod p)\] while [MATH]p | (n^2+1) [/MATH] so they're conflict, so [MATH](n^2+1,(n+2)^2+1)=1[/MATH] or [MATH]2^k[/MATH] but [MATH]a^2+1[/MATH] never divisible by 4 so [MATH](n^2+1,(n+2)^2+1)=1[/MATH] or [MATH]2[/MATH].
 
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