$\pmod 2$ tells us that either $p=2$ or $q=2$. i) $p=2$ and $q\ge 3$ $\pmod 5$ gives us $4|a$ so $a\ge 4$ and $a$ is even. We have $5q^b=(2^{\frac a2}-1)(2^{\frac a2}+1)$. Clearly, $q$ cannot divide $2^{\frac a2}-1$ and $2^{\frac a2}+1$ simultaneously. Also, both numbers are greater than $1$ as $a\ge 4$. Hence, one of them is $5$. $2^{\frac a2}-1=5\Rightarrow 2^{\frac a2}=6$, contradiction. $2^{\frac a2}+1=5\Rightarrow 2^{\frac a2}=4\Rightarrow a=4\Rightarrow q^b=2^{\frac a2}-1=3$. Hence, $\boxed{(p,q,a,b)=(2,3,4,1)}$. ii) $q=2$ and $p\ge 3$ If $a$ is odd, then $a\ge 3$ and we have $(p-1)(p^{a-1}+\cdots +p+1)=5\cdot 2^b$. Since $a$ is odd, we know that $p^{a-1}+\cdots +1$ is odd as well. Hence, $5\ge p^{a-1}+\cdots +1\ge p^2+1\ge 10$, contradiction. So $a$ is even. $\pmod 3$ tells us $p=3$. Then $\pmod 5$ tells us $4|a$. We have $5\cdot 2^b=(3^{\frac a2}-1)(3^{\frac a2}+1)$. As $4|a$, we get $2^1|| 3^{\frac a2}+1$ so $3^{\frac a2}+1\in \{2,10\}$. This gives us $p=3, a=4$ and $5\cdot 2^b=80$. So $\boxed{(p,q,a,b)=(3,2,4,4)}$.

Its easy to get that $p$ or $q$ must be even Case 1: $p=2$ $\implies 2^a= 1+5q^b \implies 2^a\equiv 1 $(mod 5) $\implies$ $a$ must be even (or more precise $a=4m $ for some $m$) $\implies 2^a\equiv 1$ mod 3 $\implies 5q^b \equiv 0$ mod 3 $\implies q=3$ Now we have that $2^a=1+5\cdot3^b$ For $b\geq2, 2^a\equiv 1 (mod 9)$ and since $ord_{9} 2=6$ we have that $a=6l$ for some $l$ This means that $2^{6l}= 1+5\cdot 3^b$ and because $(2^6)^l\equiv1$ (mod $7$) yields $7\vert 5\cdot3^b$ which is a contadiction This implies that $b=1$ and from there $a=4$ Now we investigate the second case. $p^a-1=5\cdot2^b$ Now from Zsigmondy's theorem there exists some prime divisor of $p^a-1$ that doesn't divide $p-1$ and its easy to see that that prime divisor is $5$.This means that $p$ isn't congruent to $1$ mod $5$ and $p^a\equiv 1$ (mod $5$) And this implies $a=4k$ for some $k$ Now we got that $p^{4k}-1 = 5\cdot2^b$ and if $p\neq 3$, mod $3$ gives contradiction. Now, $p=3$ and $81^{k}-1=5\cdot2^b $ Now by examining the $v_{2}$s on each side and by LTE we get: $4+v_{2}(k)=b$ or $k=2^{b-4}l >2^{b-4}$ So now $5\cdot2^b>81^{2^{b-4}} -1$ Which fails for $b>5$ (simple induction) Now we examine small cases for $b$ and see that only $b=4$ works $\implies k=1$ or $a=4$ and we finished the second case so we are done $QED$

This can be solved by Zsigmondy (although that's not a problem for a junior competition). Either $p$ or $q$ is $2$. If $q=2$ and $a>4$, then $p^a-1$ has a new prime factor which doesn't divide $p-1$, $p^2-1$, $p^3-1$, $p^4-1$. However, $5$ must divide one of those $4$ numbers because of orders. Thus, $a \leq 4$. If $a=3$, we have $(p-1)(p^2+p+1)=5 \cdot 2^b$, which implies that $p-1=2^b$ and $p^2+p+1=5$, impossible. If $a=4$, we have that $(p-1)(p+1)(p^2+1)$ is only divisible by $2$ and $5$, but this is impossible as the only common divisor of any pair of those numbers is $2$, so there are at least $3$ distinct prime factors there, unless $p=3$, in which case we get a solution. If $a=2$, either $p-1=2^{b-1}$ and $p+1=5 \cdot 2$, so $b=4$ and $p=9$, impossible, or $p+1=2^{b-1}$, $p-1=10$, which has no solution. Finally, if $q=2$, we have $2^a-1=5 \cdot p^b$. Then $a$ is divisible by $4$, but then $p=3$. However, for $a \geq 8$, there are some new prime factors other than $3$ and $5$ by Zsigmondy, so $a=4$, and this gives us another solution.

Proposed by Jason Prodromidis, Greece

Another solution without using Zsigmondy.

Your solutions are ok but it is more elegant using modulo 317. Modulo 317 yields the solution in two lines.

Using VPN is also a solution

A curious remark: all cases apart from $q = 2$ with $a$ - odd can be solved using only modular arithmetic! For $p=2$ it is easy, while for $q=2$ and even $a$ one can obtain $p=3$ from mod $3$ and then contradiction for $b\geq 5$ from moduli $32$ and $17$. (So without any factoring you can get 7/10 points lol.)

I think my solution's some parts are wrong

$(p,q,a,b)=(3,2,4,4)$($follows$ $from$ $mod$ $and$ $parity$)

$\pmod 2$ we get that either $p=2$ or $q=2$ 1) $p=2$ $2^a\equiv 1+5q^b\equiv 0\pmod 5 \implies a\equiv 0\pmod 4$ Now we have $(2^{\frac a2}-1)(2^{\frac a2}+1)=5q^b$. $GCD(2^{\frac a2}-1,2^{\frac a2}+1)=1$, thus 5 divides exactly one of the two multiples and one of the multiples is divisible by $q$. One of the multiples is 5 and the other one is $q^b$ because there can't be a multiple equal to one since $2^a-1>1$. $q^b\geq9>5+2$ contradiction because the difference of the multiples is not two. 2) $q=2$ $p^a-1=5\cdot 2^b$ $(p-1)(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)=5\cdot 2^b$ 2.1) $a$ is odd $\implies a\geq3$ and $(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)$ is odd $\implies 5\leq(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)\leq(p^2+p+1)=13$ contradiction 2.2) $a$ is even $(p^{\frac a2}-1)(p^{\frac a2}+1)=5\cdot 2^b$ $GCD(p^{\frac a2}-1,p^{\frac a2}+1)=2$, thus 5 divides exactly one of the two multiples and one of the multiples is divisible by 2 but not 4. One of the multiples is 10 and the other one 2^{b-1} because $p^a-1>2$ 2.2.1)$p^{\frac a2}-1=10$ contradiction $p\notin \mathbb{N}$ 2.2.2)$p^{\frac a2}+1=10 \implies p=3$ and $a=4$ $3^4=1+5\cdot 2^b \implies b=4$ $(a,b,p,q)=(4,4,3,2)$

mathmax12's solution (which he told me to share): If we take $\pmod{2}$, we get that either $p$ or $q$ are $2.$ If $p=2$: $2^a=1+5q^b$, taking $\pmod{5}$ gives us that $4|a$, so $5q^b=(2^{\frac{a}{2}}+1)(2^{\frac{a}{2}}-1),$ we have that one of them needs to be divisible by $5$, we get by taking cases that $(2,3,4,1)$ works. If $q=2$: By the same procedure $5\cdot 2^b=(3^{\frac{a}{2}}-1)(3^{\frac{a}{2}}+1)$, now the only solution is $(3,2,4,4).$ In conclusion $\boxed{(3,2,4,4) , (2,3,4,1)}.$

Considering the equation in mod $2$, we see that one of $p$ or $q$ must be equal to $2$. Thus, we split the problem into cases Case 1: $p=2$. The equation becomes \[2^a = 1+ 5q^b.\] Taking mod $5$ gives that \[2^a \equiv 1 \pmod{5}\]\[\implies 4 \mid a.\] Setting $a=4x$ gives \[2^{4x}=1+5q^b\]\[\implies (2^{2x}-1)(2^{2x}+1)=5q^b.\] Obviously one of these terms must equal $5$, so we are able to solve that $(p,q,a,b) = (2,3,4,1)$. Case 2: $q=2$. We plug this in to our given equation: \begin{align*} p^a-1 &= 5\cdot2^b \\ (p-1)(p^{a-1}+p^{a-2}+ \dots +1) &= 5\cdot2^b. \end{align*} Now we know that the second part does not equal $5$ and that would imply that $a$ is odd, thus $a$ has to be an even number; $a=2x$. We get \[(p^x-1)(p^x+1)=5\cdot2^b, \] so we again consider two cases: Suppose we have \begin{align*} p^x &=5\cdot2^k+1 \\ p^x &= 2^{b-k}-1 \\ 5\cdot2^k+2 &= 2^{b-k} \\ 5\cdot2^{k-1}+1 &= 2^{b-k-1} \end{align*} Clearly no solutions here. Now let the first parenthesis be $2^{b-k}$, and the second be $5\cdot2^{k}$ for $0\le k\le b.$ \begin{align*} p^x &= 5\cdot2^k-1 \\ p^x &= 2^{b-k}+1, \end{align*} giving $(p,q,a,b)=(3,2,4,4)$. Having exhausted all of the cases, our two answers are $\boxed{(2,3,4,1) \text{ and } (3,2,4,4)}$.

Wow, such a nice problem, but i may have overkilled it By taking $\pmod2$ we see that either p or q is equal to $2.$ $\textcolor{red}{CASE}$ $\textcolor{red}{1:}$ $p=2.$ We see that q is an odd number. $2^{a}\equiv 1 \pmod5\implies 4 \mid a$. Say $a=4k$ then taking $\pmod3 \implies q=3.$ The equation becomes $$2^{4k}=16^{k}=1+5.3^{b}.$$From LTE we see that $v_3(k)+1=b,$ thus $k=m.3^{b-1}$ for some integer $m\geq1.$ The equation has become $$16^{(3^{b-1})^{m}}=5.3^{b}+1.$$From the Power-Mean Inequality we have $16\geq(5.3^{b}+1)^{\frac{1}{b}}$ so now we have $$\left(5.3^{b}+1\right)^{\frac{1}{(3^{b-1})^{m}}}\geq(5.3^{b}+1)^{\frac{1}{b}}$$and since $b,m$ are positive integers this inequality only holds when $b,m=1$ meaning the only solution in this case is $\boxed{(p,q,a,b)=(2,3,4,1)}$ $\textcolor{red}{CASE}$ $\textcolor{red}{2:}$ $q=2.$ From $\pmod2$ we understand that p is odd also from $\pmod3$ we see that for $p>3$, p is an odd integer. Since $p^{a}-1=5.2^{b}$, after using Zgimondy's Theorem we see that $2 \mid p-1$ and $p^{a}-1$ has a prime divisor that $p-1$ doesnt have which is obviously 5 meaning that the only prime divisor of $p-1$ is $2.$ From the quadratic residues of $5$, we have $p\neq1,5 \pmod5\implies a=\{0,2\} \pmod5$. Thus a is an even number, and we have a contradiction. This means $p=3$. Now $p^{a}\equiv \pmod 5 \implies 4 \mid a.$ Say $a=4k.$ Using LTE we have $v_2(k)=b-4.$ By factoring we also have $$(9^{k}-1)(9^{k}+1)=5.2^{b}.$$Again using LTE leads us to $v_2(9^{k}-1)=2^{b-1}$ and $v_5(9^k+1)=1$ implies $9^k+1=10\implies k=1$, giving us the only solution in this case as $\boxed{(p,q,a,b)=(3,2,4,4)}$ $\textcolor{blue}{NOTE:}$ We didn't really have to use LTE, Zgimondy's or Power-Mean Inequality in this question, but i thought that this was kinda a cool solution.

Problem is a tad bit easy for a #3, all you do is split the parity as follows: If $a\equiv 0\pmod{2}$: If $p\neq{3}$, $p^a\equiv 1\pmod{3}$, forcing $q=3$. Now, $5\cdot{3^b}+1=p^a$. Let $a=2s$, so $p^a-1=(p^s+1)(p^s-1)=5\cdot{3^b}$. Note that both factors cannot simultaneously, since they are 2 apart from each other, and 2 is not a multiple of 3. Therefore, one of the factors is a power of 3, while the other is 5. $p^s-1>1$ unless $s=1, p=2$. Otherwise, $p^s-1=5 \Longrightarrow p^s=6$, or $p^2+1=5 \Longrightarrow p^s=4$. Obviously, only $p=2, s=2$ works after testing these cases, from which $b=1$. We get the quadruple $\boxed{(2, 3, 4, 1)}$. Now we come back to the case $s=1, p=2$, but after plugging it, in none of them produce a factor of 5. Now, we assume $p=3$. Then, $3^a-1=(3^s+1)(3^s-1)=5q^b$. Obviously, $p$ and $q$ must be of opposite parity, so that forces $q=2$. Both factors are even, and assume one of them is in the form $2^k$. The other one will be 2 away from $2^k$ so it does not divide 4 when $k>1$. Thus, one factor must be 10, and the other is $2^k=2^{b-1}$. Obviously, the only possible way to get 10 as a factor is if $s=2$, and this indeed makes the other factor $2^k$ where $k=3$. Since $k=b-1$, $b=4$. We get the quadruple $\boxed{(3, 2, 4, 4)}$. If $a\equiv 1\pmod{2}$: As established, $p$ and $q$ are different parity, so we do subcases on which one is 2. If $p=2$: $2^a\equiv 1\pmod{5} \Longrightarrow a\equiv 0\pmod{4}$ which contradicts the condition in our case. If $q=2$: $p^a-1=(p-1)(p^{a-1}+p^{a-2}\cdots+p+1)$. The second factor with the sum is odd, since there are an odd number of terms and all terms are a power of $p$ which is odd. Therefore, $p-1=2^k$ and $p^{a-1}+p^{a-2}\cdots+p+1=5$. Unfortunately, there are no sequences that satisfy this. Hence, we are done!

fun problem Notice that one of $p, q$ must be even after fixing both as odd which leads to a contradiction. We split our problem up into two cases. Case 1: $p = 2$: This means that $2^a = 1 + 5q^b$ (and $q$ is odd), which means $2^a \equiv 1\pmod5 \implies 4 \mid a$ looking at residues. Say $a = 4k$. $$2^{4k} = 1 + 5q^b \implies (2^{2k} - 1)(2^{2k} + 1) = 5q^b \implies (2^k - 1)(2^k + 1)(2^{2k} + 1) = 5q^b$$Notice that $3$ divides one of $2^k, 2^k - 1, 2^k + 1$, and since $3 \nmid 2^k$, we have $3 \mid (2^k - 1)(2^k + 1) \implies 3 \mid 5q^b \implies q = 3$. Substitute this back into the equation: $$(2^{2k} - 1)(2^{2k} + 1) = 5\cdot3^b$$Notice that $5 | \; \text{one of} \; (2^{2k} - 1)(2^{2k} + 1) \implies 2^{2k} + 1 = 5 \implies k = 1 \implies a = 4 \implies b = 1$. So our solution set for this case is simply $(4, 1, 2, 3)$. Case 2: $q = 2$: We know $p$ is odd, $p^a = 1 + 5\cdot2^b$. Take $a$ as odd: $$\implies p^a - 1 = (p - 1)(p^{a-1} + ... + p + 1) = 5\cdot2^b$$Since $a$ is odd, the right bracket is also odd, however, $a > 1 \implies p^{a - 1} + ... + 1 \geq p^2 + p + 1 \geq 13 > 5$, which means a contradiction. Therefore, $a$ is even. Say $a = 2k$. $$\implies p^{2k} - 1 = (p^k - 1)(p^k + 1) = 5\cdot2^b$$Notice that $p^k - 1, p^k + 1$ are consecutive evens, which means one only divides 2 while the other divides 4. Therefore, one of them is 10 (since $p^k = 3$, doesn't work). If $p^k - 1 = 10, p^k = 11 \implies p = 11, k = 1$. However $(11 - 1)(11 + 1)$ divides $3$, contradiction. This means $p^k + 1 = 10 \implies p = 3, k = 2 \implies a = 4 \implies b = 4$. This solution set clearly works, hence our two answers are $(4, 1, 2, 3), (4, 4, 3, 2) = (a, b, p, q)$.

The two solutions are $(p, q, a, b) = (2, 3, 4, 1)$ and $(3, 2, 4, 4)$. First, suppose $p = 2$; then we have $2^a - 1 = 5q^b$. $2^a \equiv 1 \pmod5$ so $a$ is even. Let $a = 2k$. This gives a difference of squares on the left: $(2^k + 1)(2^k - 1) = 5q^b$. Since the two factors on the LHS are odd numbers which differ by 2, they are relatively prime. Thus $5$ and $q^b$ each divide one of $\{(2^k + 1), (2^k - 1)\}$ and not the other. So the LHS must factor as $5 \cdot q^b$ or $1 \cdot 5q^b$. The first case gives $k = 2$ and $q^b = 3$, yielding the solution $(p, q, a, b) = (2, 3, 4, 1)$. The second case gives no solution. Now suppose $p > 2$. Then $p$ is odd, and checking mod 2 tells us $q^b$ is even, so $q = 2$. We will do casework on the parity of $a$. Suppose $a$ is even. Let $a = 2k$; then we have $(p^k + 1)(p^k - 1) = 5(2^b)$. Note that one of the LHS terms is $2 \pmod4$ and the other is $0 \pmod4$. So $2$ divides one term and $2^{b-1}$ divides the other. Also, $5$ divides exactly one of the terms. Thus the LHS factors as $2 \cdot 5(2^{b-1})$ or $10 \cdot 2^{b-1}$. The first case yields no solutions and the second one forces $p^k = 9$. This gives a solution $(p, q, a, b) = (3, 2, 4, 4)$. Suppose $a$ is odd; let $a = 2k+1$. We have $p^{2k+1} = 1 + 5q^b$. We have $p^{2k+1} \equiv 1 \pmod5$ which implies $p \equiv 1 \pmod5$. Since $p$ is odd, we have $p \equiv 1 \pmod{10}$. Let $p = 10m + 1$. Then: $$(10m+1)^{2k+1} - 1 = 5(2^b)$$$$(10m)((10m+1)^{2k} + (10m+1)^{2k-1} + \dots + (10m+1) + 1) = 5(2^b)$$$$(10m+1)^{2k} + (10m+1)^{2k-1} + \dots + (10m+1) + 1 = 2^{b-1}$$Taking this equation mod 2, we see that $2^{b-1}$ is odd, so $b = 1$. Then $(10m+1)^{2k+1} = 11$ has no valid solution. Therefore $(2, 3, 4, 1)$ and $(3, 2, 4, 4)$ are the only such quadruples. $\blacksquare$

Checking the parity, we have $p = 2$ or $q = 2$ $\bullet$ $p = 2$ $2^a \equiv 1 \pmod{5}$ $\Longrightarrow 4 \mid a \Longrightarrow$ $(2^{\frac{a}{2}} - 1)(2^{\frac{a}{2}} + 1) = 5q^b$ so, $2^{\frac{a}{2}} + 1 = 5$ $\Longrightarrow a = 4 \Longrightarrow$ $q=3$ and $b=1$ $\boxed{\therefore (p, q, a, b) = (2, 3, 4, 1)}$ $\bullet$ $q = 2$ If $a$ is odd, notice that: $p^a \equiv 1 \pmod{5\cdot2^b}$ hence, $\begin{cases} ord_{5\cdot2^b}(p) \mid a \\ ord_{5\cdot2^b}(p) \mid \phi(5\cdot2^b) = 2^{b+1} \end{cases}$ $\Longrightarrow ord_{5\cdot2^b}(p) = 1 \Longrightarrow$ $p = (5\cdot2^b)k + 1 \geq$ $5\cdot2^b + 1 = p^a$$ABS!$ thus, $a$ is even and $(p^{\frac{a}{2}} - 1)(p^{\frac{a}{2}} + 1) = 5\cdot2^b$ $\Longrightarrow$ $\begin{cases}p^{\frac{a}{2}} - 1 = 10 & p = 11, a = 2 \\ p^{\frac{a}{2}} + 1 = 10 & p = 3, a = 4\end{cases}$ The first case gives us $2^b = 24 ABS!$, and, the second case gives us $b = 4$ $\boxed{\therefore (p, q, a , b) = (3, 2, 4, 4)}$

I may have overkilled this one Taking mod 2, we have either \(p\) or \(q\) even. So, we may split this problem into two cases. Case 1: \(p = 2\) We have \(2^a = 1 + 5q^b\). Since \(2^a \equiv 1 \mod{5}\), \(a \equiv 0 \mod{4} \implies 2^a \equiv 1 \mod{3}\). Therefore, \(3 \mid 5q^b \implies q = 3\). Observe that \(v_3(2^a - 1) = 1 + v_3(a)\). If \(3 \mid a\), we have \(2^a \equiv 1 \mod{7} \implies q = 7\), which is impossible. Therefore, \(v_3(5q^b) = 1 \implies b = 1\). Substituting this back, we have \(a = 4\). So, our solution in this case is \((p, q, a, b) = (2, 3, 4, 1)\) \(\square\) Case 2: \(q = 2\) We have \(1 + 5 \cdot 2^b = p^a\). We shall divide this one into two subcases. Subcase 1: \(b\) is even This implies \(5 \cdot 2^b + 1 \equiv 0 \mod{3} \implies p = 3\), so we have \(1 + 5 \cdot 2^b = 3^a\). \(b = 1\) fails, so \(3^a \equiv 1 \mod{4}\), which means \(a\) is even. The equation rearranges to \(3^a - 1 = 5 \cdot 2^b \implies v_2(3^a - 1) = 2 + v_2(a) = b\), so \(a \geq 2^{b - 2}\). Notice that for \(b \geq 5\), we have \(3^{2^{b - 2}} > 1 + 5 \cdot 2^b\) (induct ). \(b = 2\) fails (it doesn't give a power of 3), and \(b = 4\) gives \(a = 4\). So, another solution is \((p, q, a, b) = (3, 2, 4, 4)\). Subcase 2: \(b\) is odd This means \(1 + 5 \cdot 2^{b} \equiv 2 \mod{3} \implies p \equiv 2 \mod{3}\), \(a \equiv 1 \mod{2}\). Notice that \(p \equiv 1 \mod{2}\), \(p^a \equiv 1 \mod{4} \implies p \equiv 1 \mod{4}\). Assume \(p \equiv 1 \mod{2^k}\) for \(k \leq b - 1\). We claim \(p \equiv 1 \mod{2^{k + 1}}\) as well. Assume that it's not. This means \(p \equiv 2^k + 1 \mod{2^{k + 1}}\). This means \(p^2 \equiv 1 \mod{2^{k + 1}} \implies p^{2x} \equiv 1 \mod{2^{k + 1}} \implies p^{2x + 1} \equiv 2^k + 1 \mod{2^{k + 1}}\), which is a contradiction since \(a\) is odd as well. Therefore, \(p \equiv 1 \mod{2^{k + 1}}\), and \(p \equiv 1 \mod{2^b} \implies p \geq 2^b + 1\). This means \(p^a \geq (2^b + 1)^a\). However, since \(a \geq 3\), we have \((2^b + 1)^a > 5 \cdot (2^b + 1)\) for all \(b\). So, there are no solutions in this case. Therefore, our solutions are only \((p, q, a, b) = (2, 3, 4, 1), (3, 2, 4, 4)\). $\square$

$\boxed{ANSWER:(p,q,a,b)=(2,3,4,1)(3,2,4,4)}$ claim 1:$p,q=2$ proof:$p\equiv 1+q\pmod{2}\implies p,q=2$ case 1:$q=2$ $p^{a}=1+5\cdot 2^{b}$ $p^a-1=(p-1)(p^{a-1}+p^{a-2}\cdots+p+1)\implies a=2a_1$ $(p^{a_1}-1)(p^{a_1}+1)=5\cdot 2^b$ case 1.1:$p^{a_1}-1=2\implies p=3,a=2\implies contradiction$ case 1.2:$p^{a_1}+1=10\implies p=3,a=4\implies b=4$ case 2:$p=2$ $2^a=1+5\cdot q^b$ $\pmod{5}\implies a=4a_1$ $(2^{2a_1}-1)(2^{2a_1}+1)=5\cdot q^b$ $gcd=1\implies$ one of them is $5$ case 2.1:$2^{2a_1}-1=5\implies contradiction$ case 2.2:$2^{2a_1}+1=5\implies a=4\implies q=3,b=1$