Question

Show that if${\mathrm{n}}^2+=1$is a perfect square, wherenis an integer, then nis even.

Short answer

n is even.

Step-by-step solution

Step 1

Given: $n^2+1$ is a perfect square

Proof: n is even

PROOF BY CONTRADICTION

Let us assume that is odd

First part Since n is odd, then there exists an integer k such that:

n = 2k + 1

Determine the square of each side of the previous equation:

$n^2={\left(2k+1\right)}^2$

Use the property $(a+b{)}^2=a^2+2ab+b^2$

$n^2=4k^2+4k+1$

Add 1 to each side of the previous equation:

$n^2+1=4k^2+4k+2$

We then note $n^2+1=2\left(mod4\right)$.

Second part Since n is odd, $n^2$is also odd thus $n^2$+ 1 is even. If $n^2+1$is even and a perfect square, then there exists an even integer m such that

role="math" localid="1668506420753" $n^2+1=m^2$

We then note that there exists an integer such that m = 2 p

$n^2+1=m^2={\left(2p\right)}^2=4p^2$

We then note that ${\mathrm{n}}^2+1=0(\mathrm{mod}4)$

ConclusionWe have then obtained a contradiction, because it is not possible that $n^2+1\equiv 2(mod4)\text{and}{n}^2+1\equiv 0(mod4)$. Our assumption that is odd then false and thus n is even.