# About Fermat numbers

I am self studying elementary number theory by David Burton and got stuck on this problem in exercise $$11.4$$ of chapter - numbers of special form.

Question is:

for $$n> 0$$ prove that $$F_n$$ ( Fermat number $$n$$) is never a triangular number.

I tried to attempt it by assuming it to be triangular and obtaining contradiction in quadratic equation thus formed but was struck. Please help.

Let's just try to solve it directly. We want $$2^{2^n}+1=\frac {k(k+1)}2\implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$\sqrt {1+8(2^{2^n}+1)}\in \mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2\implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $$m-3,m+3$$ powers of $$2$$ but since $$m-3, m+3$$ differ by $$6$$ the only solutions would be very small. Indeed the only powers of $$2$$ that differ by $$6$$ are $$(2,8)$$. It is easy to show that $$n=0, k=2,m=5$$ is the only small solution and the above shows that there are no large ones, so we are done.