Give it a try

Number Theory Level 2

If $m$ and $n$ are positive integers, which of these cannot be equal to $m^2$ ?

n^2 + 5 n^5 - 4 n^3 n^3 + 1

2 solutions

Lee Wall
Apr 5, 2014

In each case, we attempt to construct a counterexample. For $m^2 = n^3+1$ , we have the unique solution $(m,n) = (3,2)$ . For $m^2 = n^3$ , we can construct an infinite number of counterexamples of the form $a^6 = (a^2)^3 = (a^3)^2$ . For $m^2 = n^2+5$ , a bit of rearrangement yields $(m-n)(m+n) = 5$ , which has the solution $(m, n) = (3, 2)$ . This leaves $n^5-4$ as the answer.

But, how do you prove that there is no integer solution for $n^5-4 = m^2$ ?

Janardhanan Sivaramakrishnan - 6 years, 5 months ago

I did it in a similar way.....CHEERS!!!

Vighnesh Raut - 7 years, 1 month ago

same way, bro!! mcq makes life easier

Aareyan Manzoor - 6 years, 6 months ago

Check again, Evan. (1,1) is a solution of m^2 = 5*n - 4, but not a solution of m^2 = n^5 - 4.

Richard Desper - 4 years, 7 months ago

n = 81, m = 59049 m^2 = n^5 - 4

mindany2 mindany2 - 4 years, 6 months ago

Presenting a counter example m^2=n^5 - 4 We have a solution set of (1,1)

Evan Glori - 6 years, 2 months ago

Shivam Jadhav
Apr 5, 2014

$8^{2} = 4^{3}, 3^{2} = 2^{2} + 5 = 2^{3} + 1,$ to prove (A) take $m = 11k + p, n = 11k + q.$

Give it a try

2 solutions

0 pending reports