$ (2^{2m + 1})^2 + 1 = (2^{2m + 1} + 1)^2 - 2^{2m + 2} = (2^{2m + 1} + 2^{m + 1} + 1)(2^{2m + 1} - 2^{m + 1} + 1)$. Thus, we have cases: 1)$ 2^{2m + 1} - 2^{m + 1} + 1 = 1$. 2)$ 2^{2m + 1} - 2^{m + 1} + 1 = p^a, 2^{2m + 1} + 2^{m + 1} + 1 = q^b$.

Friend ,please explain how that solves the problem..I got till this and also the shorter of those numbers is a power of $ 5$ ....what about this case ,you left this ..... $ 2^{4m + 2} + 1 = p^{a}q^{b}$ where $ p^{a} = 5^{a} = 2^{2m + 1} + 1 - 2^{m +1}$ . I dont think the first case of yours is possible .....not too sure though ......

that's baltic way 2002: go to (in fact i mentioned a special case of the problem but an explanation is given in the last topic) http://www.mathlinks.ro/viewtopic.php?p=1578991#1578991

I have not read your link shoki, so i dont know if the following ideas are the same as those ones.. $ 2^{4m+2}+1$ is in the form $ x^4+4y^4$ (Sophie-German identity), so it can be factorized as $ ((2^m)^2+(2^m-1)^2)((2^m)^2+(2^m+1)^2)$. Since the two factor are coprime,both odd and greater than $ 1$ then we are lead to solve the system $ p^a=(2^m)^2+(2^m+1)^2>(2^m)^2+(2^m-1)^2=q^b$, for some distinct prime $ p,q$ (for istance $ 4 \mid \text{gcd}(p-1,q-1)$ since $ 1=(\frac{-1}{p})=(\frac{-1}{q})$) and $ a,b$ are positive integers. Now we have three consecutive quadratic residue of (x-1)²,x²,(x+1)², where x² is non zero for every mod p>2. Modulo 5 we see that if x² is a fixed r in {-1,1} then at least one between (x-1)² and (x+1)² is -r. It means that (p-5)(q-5)=0. So, in other words, it sufficies to find all solutions of the equation $ (2^m)^2+(2^m+\ell)^2=5^a$, where $ l \in \{-1,1\} \text{ and } a \in \mathbb{N}_0$. If $ m>1$ we can see modulo $ 8$ that $ 2 \mid a$, so it exists $ b \in \mathbb{N}_0$ such that $ a=2b$, so $ 2^{2m}=(5^b+2^m+\ell)(5^b-2^m-\ell)$. Now $ \text{gcd}(5^b+2^m+\ell,5^b-2^m-\ell)$ $ =\text{gcd}(5^b+2^m+\ell,2 \cdot 5^b)$ and must be 2, so it sufficies to solve $ 2^{2m-1}=5^b+2^m+\ell \text{ and } 2=5^b-2^m-\ell$ $ \implies 5^b=2^m+\ell+2=2^{2m-1}-2^m-\ell$. It implies that $ 2^{2m-1} = 2^{m+1}+2\ell +2 \le 2^{m+1}+4$ that is false if $ m>2$. So, unique solutions are $ m \in \{1,2\}$ that is verified by hand.[]

@bbyopa Friend ,can you tell me the reason why you have assumed $ (2^{m})^{2}+(2^{m}+\ell)^{2}=5^{a}$, I mean cant we directly tell that $ (2^{m})^{2} + (2^{m} - 1 )^{2} =5^{a}$ because $ 5$ always divides $ 2^{4m+2} +1$

In principle i have complicated my life since i have not seen that $ 4^{2x + 1} = -1 \text{ in } \mathbb{F}_5$, on the other hand both ways work, so what is the problem? Ps. But why have you assumed $ \ell = - 1$ in your statement?

er.......I never assumed $ \ell = 1$ ,infact I never even introduced a $ \ell$ , I just did this We have $ (2^{2m + 1} - 2^{m + 1} + 1)(2^{2m + 1} + 2^{m + 1} + 1) = 5^{a}q^{b}$ Since $ \gcd( 2^{2m + 1} - 2^{m + 1} + 1 , 2^{2m + 1} + 2^{m + 1} + 1) = 1$ and also $ q > 5$ and $ 2^{2m + 1} - 2^{m + 1} + 1 < 2^{2m + 1} + 2^{m + 1} + 1$ ,we have $ 2^{2m + 1} - 2^{m + 1} + 1 = 5^{a}$ , from this I got a different solution . P.S -Friend can you tell me what is $ \mathbb{F}_{5}$

Well here is my solution ,sorry if this is almost the same as yours . We observe that $ m = 1$ and $ m = 0$ form a solution ,so assume $ m \neq 1,0$ I put forth the following lemma and assumptions. $ \text{Lemma 1}$ $ 2^{4m + 2} + 1 \equiv 0\bmod{5}$ ,so definitely $ 5$ is one of the prime dividing $ 2^{4m + 2} + 1$ $ \text{Assumption 1}$ $ 2^{4m + 2} + 1 = 5^{\alpha}q^{\beta}$ where $ q$ is a prime and $ \alpha ,\beta \in \mathbb{N}$ $ \text{Assumption - 2}$ $ 2^{4m + 2} + 1 = 5^{\alpha}$ Let us work with Assumption 1 We get $ 2^{4m + 2} + 1 = (2^{2m + 1} + 1 - 2^{m + 1})(2^{2m + 1} + 1 + 2^{m + 1}) = 5^{\alpha}q^{\beta}$ From my previous post I conclude that $ (2^{2m + 1} + 1 - 2^{m + 1}) = 5^{\alpha}$ $ \text{Lemma - 2}$ $ \alpha$ is even . The proof can be directly got by factorising the R.H.S Let $ \alpha = 2k$ where $ k \in \mathbb{N}$ $ (2^{2m + 1} + 1 - 2^{m + 1}) = (2^{m})^{2} + (2^{m} - 1)^{2} = 5^{2k}$ $ (2^{m})^{2} = (5^{k} + 2^{m} - 1)(5^{k} - 2^{m} + 1)$ $ \implies (5^{k} + 2^{m} - 1) = 2^{a},(5^{k} - 2^{m} + 1) = 2^{b}$ where $ a,b \in \mathbb{N}$ and $ a + b = 2m$ Now I am going to consider $ \gcd(5^{k} + 2^{m} - 1 , 5^{k} - 2^{m} + 1)$ If $ m \ge 2$ ,we have $ 4 \mid 5^{k} + 2^{m} - 1$ but $ 4 \nmid 5^{k} - 2^{m} + 1$ Hence $ \gcd(5^{k} + 2^{m} - 1 , 5^{k} - 2^{m} + 1) = 2$ $ \ldots \boxed{1}$ $ a \ge 2$ then from $ \boxed{1}$ we have $ b = 1$ So we get $ 5^{k} - 2^{m} + 1 = 2$ $ 5^{k} = 2^{m} + 1$ Since $ 2^{m} \equiv 4\bmod{5}$ $ \implies m\equiv 2\bmod{4}$ ,let $ m = 4l + 2$ where $ l \in \mathbb{N}_{0}$ $ 5^{k} = (2^{2l + 1} + 1 - 2^{l + 1})(2^{2l + 1} + 1 + 2^{l + 1})$ As $ \gcd(2^{2l + 1} + 1 - 2^{l + 1}, 2^{2l + 1} + 1 + 2^{l + 1})$ and $ 2^{2l + 1} + 1 - 2^{l + 1} < 2^{2l + 1} + 1 + 2^{l + 1}$ ,we have $ 2^{2l + 1} + 1 - 2^{l + 1} = 1$ and$ 2^{2l + 1} + 1 + 2^{l + 1} = 5^{k}$ From the first equation we get $ l = 0$ and $ m = 2$ So the only solutions are $ m\in\{0,1,2\}$ $ \blacksquare$ Sorry if my solution and yours are same .

Yes, the solutions are pratically the same but don't worry .. srinath.r wrote: P.S -Friend can you tell me what is $ \mathbb{F}_{5}$ Sure, $ \mathbb{F}_5$ is the finite field with $ 5$ element and (for our aim!) i could write also $ \mathbb{Z}/5\mathbb{Z}$ or $ \text{mod}5$. srinath.r wrote: ..We have $ (2^{2m + 1} - 2^{m + 1} + 1)(2^{2m + 1} + 2^{m + 1} + 1) = 5^{a}q^{b}$. Since [...] $ q > 5$ and $ 2^{2m + 1} - 2^{m + 1} + 1 < 2^{2m + 1} + 2^{m + 1} + 1$ ,we have $ 2^{2m + 1} - 2^{m + 1} + 1 = 5^{a}$ But are you agree that we can have $ q^b < 5^a$?

er........That neednt be true always ..anyway I just have to include another case -$ 2^{2m+1} -2^{m+1} + 1 = q^{\beta}$ and proceeding in the previous way ...You can solve

Also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=360424