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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.

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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.

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  • $\begingroup$ Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that. $\endgroup$ – AMDE Dec 3 '18 at 3:18

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