# Math Help - perfect squares

1. ## perfect squares

A few problems:

1) Show that for positive integers n, n^4 + 3n^2 + 1 is never a perfect square.

2) Let f(n) = 2n^2 + 14n + 25. We see that f(0) = 25 = 5^2. Find two positive integers n such that f(n) is a perfect square.

3) Find all pairs of positive integers (x,y) such that both x^2 + 3y and y^2 + 3x are perfect squares.

2. 1) $n^4+3n^2+1=m^2$

$\implies\ n^4+3n^2+(1-m^2)=0$

$\implies\ n^2=\frac{-3\pm\sqrt{3^2-4(1-m^2)}}{2}=\frac{-3\pm\sqrt{5+4m^2}}{2}$

$5+4m^2$ is only a perfect square for $m^2=1$, which would only give $n=0$.

3. Originally Posted by scipa
A few problems:

1) Show that for positive integers n, n^4 + 3n^2 + 1 is never a perfect square.

Suppose that there exists a positive integer $n$ such that: $n^4+3n^2+1$ is a perfect square $k^2$.

Then clearly $k>n^2+1$, since otherwise $k^2$ would be less than $n^4+3n^2+1$.
Also $k, since otherwise $k^2$ would be greater than $n^4+3n^2+1$.

But there are no integers $k$, such that $n^2+1.

RonL

4. Originally Posted by scipa
A few problems:

1) Show that for positive integers n, n^4 + 3n^2 + 1 is never a perfect square.

2) Let f(n) = 2n^2 + 14n + 25. We see that f(0) = 25 = 5^2. Find two positive integers n such that f(n) is a perfect square.

3) Find all pairs of positive integers (x,y) such that both x^2 + 3y and y^2 + 3x are perfect squares.
For 2) it can be show that for each integer $m$, $n=1/4\, \left( 1+\sqrt {2} \right) ^{2\,m-1}+1/4\, \left( 1-\sqrt {2}
\right) ^{2\,m-1}-7/2$
is an integer such that f(n) = 2n^2 + 14n + 25 is a square.

5. Originally Posted by scipa
A few problems:

1) Show that for positive integers n, n^4 + 3n^2 + 1 is never a perfect square.

2) Let f(n) = 2n^2 + 14n + 25. We see that f(0) = 25 = 5^2. Find two positive integers n such that f(n) is a perfect square.

3) Find all pairs of positive integers (x,y) such that both x^2 + 3y and y^2 + 3x are perfect squares.
For problem 3), (1,1), (11,16), (16,11) are solutions. I would be interested to learn how to prove there are no more.