The left hand side is even, so $ a^4+b^2$ is even, and $ a,b$ has the same parity. However, if $ a,b$ are odd, we have that $ 4$ divides $ 7^a-3^b$, but $ 4$ doesn't divide $ a^4+b^2$. So $ a,b$ are even, and let's say $ a=2a_1,b=2b_1$. You'll get $ (7^{a_1}-3^{b_1})(7^{a_1}+3^{b_1})$ divides $ 4(4a_1^4+b_1^2)$ If $ b_1$ is odd, then $ 8$ divides $ (7^{a_1}-3^{b_1})(7^{a_1}+3^{b_1})$ but not $ 4(4a_1^4+b_1^2)$, so $ b_1$ is even. Say $ b_1=2c_1$. You'll get $ (7^{a_1}-3^{2c_1})(7^{a_1}+3^{2c_1})$ divides $ 16(a_1^4+c_1^2)$ So $ |(7^{a_1}-3^{2c_1})(7^{a_1}+3^{2c_1})|\leq |16(a_1^4+c_1^2)|$ But $ |(7^{a_1}-3^{2c_1})\geq 2$ (because both terms are even), and so $ 2(7^{a_1}+3^{2c_1})\leq 16(a_1^4+c_1^2)$. But for $ a_1=1,2,3$ we easily get $ c_1=1,2$, and for $ a_1\geq 4$, we have that $ 2(7^{a_1})>16a_1^4$. Also, $ 2(3^{2c_1})>16c_1^2$ for all $ c_1\in \mathbb{N}$. So it is enough to check $ (a_1,c_1)=(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$. We'll get $ (a_1,c_1)=1,1$, or $ (a,b)=(2,4)$ as the only solution. By the way, Dida Drogbier, are you from Chelsea or AC Milan?

dig, dig, dig alternative solution anybody?

This is my alternative solution. The left hand side is even, so $ a^4+b^2$ is even, and $ a,b$ has the same parity. However, if $ a,b$ are odd, we have that $ 4$ divides $ 7^a-3^b$, but $ 4$ doesn't divide $ a^4+b^2$. So $ a,b$ are even, and let's say $ a=2x,b=2y$. You'll get $ (7^{x}-3^{y})(7^{x}+3^{y})$ divides $ 4(4x^4+y^2)$ If $ y$ is odd, then $ 8$ divides $ (7^{x}-3^{y})(7^{x}+3^{y})$ but not $ 4(4x^4+y^2)$, so $ y$ is even. Say $ y=2z$. You'll get $ (7^{x}-3^{2z})(7^{x}+3^{2z})$ divides $ 16(x^4+z^2)$ So $ |(7^{x}-3^{2z})(7^{x}+3^{2z})|\leq |16(x^4+z^2)|$ But $ |(7^{x}-3^{2z})\geq 2$ (because both terms are even), and so $ 2(7^{x}+3^{2z})\leq 16(x^4+z^2)$. But for $ x=1,2,3$ we easily get $ z=1,2$, and for $ x\geq 4$, we have that $ 2(7^{x})>16x^4$. Also, $ 2(3^{2z})>16z^2$ for all $ z\in \mathbb{N}$. So it is enough to check $ (x,z)=(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$. We'll get $ (x,z)=1,1$, or $ (a,b)=(2,4)$ as the only solution.

tom: you shouldn't give solution to shortlist problem before IMO. you should talk to me (phone number below)

I believe that proving $b \equiv 0 \pmod{4}$ is unnecessary, the solution in that case is a bit more messy, but similar. The motivation for this solution is using size issues in divisibility, similar to the above posts.

AMN300 wrote: $4d^2-3^d >0 \implies d=1$ why doesn't $4d^2-3^d >0$ imply $d=1,2,or 3 $ ?

It feels nice to get back to NT. Here's my solution that uses the same idea of bounding. Proof: Notice that clearly $7^a-3^b \equiv 0 \pmod 2$. Since $7,3$ have the same parity so this forces that either $(a,b)$ are both odd or are both even. We'll firstly work with the case when $(a,b)$ are both odd. Let $(a,b)=(2k+1,2l+1)$. Notice that $$7^a-3^b \equiv 7^{2k+1}-3^{2l+1} \equiv 0 \pmod 4$$But $a^4+b^2 \equiv (2k+1)^4 +(2l+1)^2 \equiv 2 \pmod 4$ a contradiction. Now let $(a,b)=(2k,2l)$. Clearly we must have that $(7^k+3^l)(7^k-3^l) \mid 16k^4+4l^2$. But since $$2(7^k+3^l) \leq |(7^k+3^l)(7^k-3^l)| \leq 16k^4+4l^2$$so we can estabilish that $7^k+3^l \leq 8k^4+2l^2$. This is equivalent to $3^l-2l^2 \leq 8k^4-7^k$. But notice that from induction $\forall l \geq 1$ we have that $3^l>2l^2$ and hence we have $0 \leq 3^l-2l^2 \leq 8k^4-7^k$ . But notice that $7^k \geq 8k^4$ $\forall k \geq 4$ hence we have to check for $k \leq 3$. Now notice that if $k=3$ then we have that $343+3^l \leq 648+2l^2 \implies 3^l-2l^2 \leq 305$. But notice that $3^l -2l^2 > 305$ for all $l\geq 6$ hence we have to check till $l=6$. But clearly none of the solution sets work. $\square$. Now if $k=2$ then we have $3^l-2l^2 \leq 128-49=79$ . But again notice that we that $3^l-2l^2 > 78$ for $l\geq 5$. Hence we check all the way uptil $5$. But clearly none of the solution sets work again $\square$. Now if $k=1$ then notice that we have $3^l-2l^2 \leq 1$ but notice that this is true only when $l \leq 2$ . So we check till $l=2$ and we have a solution set that woks which is $(k,l)=(1,2)$ or that $\boxed{(a,b)=(2,4)}$ $\blacksquare$.

tom_damrong wrote: This is my alternative solution. Ummm... You gave the exact solution (quite literally) as the one two posts above. How is yours an "alternative" solution?

Surprisingly, if you push hard enough this problem just falls. We claim only $(2,4)$ works, and we can easily verify that it does. Note that $2\mid 7^a-3^b,$ so $2\mid a^4+b^2,$ implying $a\equiv b\pmod{2}.$ This implies that $7^a\equiv 3^b\pmod {4},$ so $4\mid a^4+b^2,$ implying that $2\mid a$ and $2\mid b.$ Say that $a=2x$ and $b=2k.$ Then note $8\mid 49^{x}-9^k,$ so we must have $8\mid 4(4x^2+k^2),$ implying that $2\mid k.$ Let $k=2y,$ then we want to solve $(7^x-9^y)(7^x+9^y)\mid 16(x^4+y^2).$ Now we just bound. Note that $2(7^x+9^y)\leq |(7^x-9^y)(7^x+9^y)|\leq 16(x^4+y^2),$ implying $7^x+9^y\leq 8(x^4+y^2).$ Note that for all $x\geq 4,$ we have $7^x> 8x^4,$ and for all positive integers $y,$ $9^y>8y^2.$ So the only possible solutions are natural numbers $x\leq 3.$ In the case of $x=1,$ note that we must have \[(7-9^y)(7+9^y)\mid 16(1+y^2),\]or $81^y-49\leq 16y^2+16.$ Note this fails for any $y\geq 2,$ and that $y=1$ yields equality, so $(1,1)$ is the only solution for $x=1.$ In the case of $x=2,$ note that we must have \[(49-9^y)(49+9^y)\mid 16(16+y^2).\]We can easily verify that $y=1$ fails. For $y\geq 2,$ we want to ensure $81^y-7^4\leq 16y^2+16\cdot 16,$ which fails for all $y\geq 2.$ In the case of $x=3,$ note that we must have \[(343-9^y)(343+9^y)\mid 16(81+y^2).\]We can easily verify that $y=1$ and $y=2$ fail by size reasons, noting $343-9^y<16$ and $343+9^y>81+y^2$ for $y=1,2.$ For $y\geq 3,$ we want to ensure that $81^y-7^6\leq 16y^2+16\cdot 81,$ which fails for all $y\geq 3.$ So the only solution in $(x,y)$ is $(1,1),$ corresponding to $(2,4).$

If $a$ and $b$ don't have the same parity, then $7^a-3^b$ is even and $a^4+b^2$ is odd, absurd. Otherwise, $4\mid 7^a-3^b\mid a^4+b^2$ so $a$ and $b$ are both even. Let $a=2m,b=2n$. Then $(7^m-3^n)(7^m+3^n)\mid 16m^4+4n^2$. Now, if $m$ and $n$ are the same parity, the LHS is divisible by $8$ and otherwise it is still divisible by $8$, so either way $2\mid n$ and thus we can let $n=2k$ so $(7^m-9^k)(7^m+9^k)\mid 16(m^4+k^2)$. Now, we use the crowbar known as "size": it is clear that the LHS's absolute value is at least $2(7^m+9^k)$, so at least one of $8k^2\ge 9^k$ and $8m^4\ge 7^m$ occurs. The former can never occur and the latter can only occur for $m\le 3$. Now, at $m=3$ we get $(343-9^k)(343+9^k)\mid 16(81+k^2)$. But the value of $k$ that minimizes $|343-9^k|$ is $262$, which is bad because $16\cdot81/262<343$. If we have $m=2$, we get $(49-9^k)(49+9^k)\mid 16(16+k^2)$. Similarly to before, note $|49-9^k|$ is minimized at $32$, so since $16\cdot16/32=8<49$, this is bad. So finally $m=1$ and $49-81^k=(7-9^k)(7+9^k)\mid 16(1+k^2)$. Equivalently, $0<81^k-49\mid 16(1+k^2)$. Equality holds exactly at $k=1$. Moreover, $\log(81)\cdot 81>32$ and $\log(81)^2\cdot81>32$, so by differentiation the result follows: the divisibility holds only when $(a,b)=(2,4)$, which can easily be checked to be true.

Quite straightforward. Note that $2\mid 7^a-3^b\implies 2\mid a^4+b^2$. Hence $a$ and $b$ have the same parity. If $a$ and $b$ are both odd, then $7^a\equiv (-1)^a\equiv -1\equiv (-1)^b\equiv 3^b$ (mod $4$) $\implies 4\mid 7^a-3^b$, which is contradictory to the congruence $7^a+3^b\equiv 2$ (mod $4$). Thus $a$ and $b$ are both even,let $a=2m,b=2n$. The divisibility condition becomes $7^{2m}-3^{2n}\mid 16m^4+4n^2\implies (7^m+3^n)(7^m-3^n)\mid 16m^4+4n^2$. Since $2\mid 7^m-3^n$,we have $7^m+3^n\le 8m^4+2n^2$. Then we can check case by case and get the unique solution $(a,b)=(2,4)$.

This took me way too long (expected to be cuter ). This solution is essentially the same as m.essein's, but posted for fun. We have by modulo $4$ reasons, that $a$ and $b$ are even. Taking also modulo $8$, we get that $a=2x$ and $b=4y$. So essentially, we reduce our problem to the following: $$(7^x-9^y)(7^x+9^y)\mid 16(x^4+y^2).$$Thus, $7^x+9^y\leq 8(x^4+y^2)$ as $2\mid 7^x-9^y\implies 2\leq \vert 7^x-9^y \vert$ . Note that $7^x>8x^4$ for all $x>3$ and $9^y> 8y^2$ for all $y>0$. Therefore we must only consider values $x=1,2,3$. Note that if $x\leq 3$, then $y=1$ or $2$. Thus, we only need to consider pairs $(x,y)=(1,1),(1,2),(2,1),(2,2),(3,1),(3,2)$. Only solution is $(x,y)=(1,1)$. Thus, we obtain our only solution pair $\boxed{(a,b)=(2,4)}$.

Notice that $7^a-3^b$ must be even, as both terms are odd. This implies that $a^4+b^2$ is even, so $a$ and $b$ have the same parity. However, if $a$ and $b$ were both odd, then we would have $$LHS\equiv (-1)^a-(-1)^b \equiv (-1)-(-1)\equiv 0 \pmod 4,$$as $a$ and $b$ are odd, and $$RHS \equiv 1+1\equiv 2 \pmod 4$$as an odd perfect square must be $1\pmod 4$, which is a contradiction. Therefore $a$ and $b$ are both even. This means that if we let $a=2c$ and $b=2d$, then $$49^c-9^d | 16c^4+4d^2.$$Then, since $49^c-9^d=(7^c+3^d)(7^c-3^d)$, we must also have $$7^c+3^d | 16c^4+4d^2,$$in particular $7^c+3^d \le 16c^4+4d^2$. We must have $(7^c-16c^4)+(3^d-4d^2)<0$. The minimum of $7^c-16c^4$ is $-1695$ at $c=4$, and $3^d-4d^2>1695$ for $d\ge 7$, so $d<7$. Similarly, the minimum of $3^d-4d^2$ is $-9$ at $d=3$, and $7^c-16c^4>9$ for $c\ge 5$, so $c<5$. Checking the remaining cases, we get that $c=1, d=2$ is the only one that works, which corresponds to $(a,b)=\boxed{(2,4)}$.

Solved with L567 using the Catalan Conjecture (well technically not but yeah) - posting for storage. Notice that $2 \mid 7^a-3^b \mid a^4+b^2$ meaning that $a \equiv b \pmod{2}$ and therefore $4 \mid 7^a-3^b \mid a^4+b^2$ and consequently $2 \mid a,b$, writing $a = 2x,b=2y$, we have $$(7^x-3^y)(7^x+3^y)=7^{2x}-3^{2y} \mid 4(4x^4+y^2)$$Notice that $\mid 7^x-3^y \mid \geq 2$ (by the Catalan Conjecture) therefore $$7^x+3^y \leq 8x^4+2y^2$$Notice that $7^x > 8x^4$ for all $x \geq 4$ and $3^y > 2y^2$ for all $y \geq 3$. Checking all the cases for $x \in \{1,2,3\}$ and $y \in \{1,2\}$ gives us the only pair $(a,b) = (2,4)$.

Solution from Twitch Solves ISL: The only answer is $(a,b) = (2,4)$, which works as $7^2 - 3^4 = -32$ divides $2^4 + 4^2 = 32$. To show this is the only solution, we note the following: The left-hand side is always even, which implies $a$ and $b$ have the same parity. This in turn implies the left-hand side is $0 \bmod 4$. Hence $a^4+b^2 \equiv 0 \pmod 4$ which means $a$ and $b$ are even. This means that $7^a -3^b \equiv 0 \pmod 8$, which further means $b \equiv 0 \pmod 4$. This means we may let $a = 2x$ and $b = 2y$ with $y$ even to obtain \[ \left( 7^x - 3^y \right)\left( 7^x + 3^y \right) \mid (2x)^4 + (2y)^2. \]To reduce to finite casework we contend: Claim: $2 \cdot 3^a > 16a^4 + 4a^2$ once $a \ge 10$. Proof. $3^{10} = 243^2 = 2 \cdot59049 = 177147 > 160400$. $\blacksquare$ Hence we are done if $\max(x,y) \ge 10$. Next Claim: $7^x > 16x^4 + 4 \cdot 10^2$ for $x \ge 6$ Proof. $7^6 = 343^2 > 100000 > 16 \cdot 6^4 + 400$. $\blacksquare$ So we may assume $x \le 5$ as well. This gives us the following cases. In all situations, note that $(2 \cdot 5)^4 + (2 \cdot 10)^2 = 10400$ and consequently if $(7^x-3^y)(7^x+3^y) > 10400$ there is nothing to verify. These cases are marked as ``big'' in the table. \[ \begin{array}{rr|ccccc} && 7 & 49 & 343 & 2401 & 16807 \\ && x=1 & x=2 & x=3 & x=4 & x=5 \\ \hline 9 &y=2 & \text{OK} & 40 \cdot 58 \nmid 272 & \text{big} & \text{big} & \text{big} \\ 81 &y=4 & 74 \cdot 88 \nmid 48 & 32 \cdot 130 \nmid 320 & \text{big} & \text{big} & \text{big} \\ 729 &y=6 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ 6561 &y=8 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ \end{array} \]

LHS is even, so RHS is also even, so $a$ and $b$ have same parity. This implies $7^a-3^b\equiv 0\pmod 4$, so $a$ and $b$ are even, which implies $7^a-3^b\equiv 0\pmod 8$, so $b\equiv 0\pmod 4$. Set $a=2x$ and $b=2y$ where $y$ is even. So $(7^x-3^y)(7^x+3^y)\mid 16x^4+4y^2$. We note that if $y>3$, then $3^y>4y^2$. If $x>4$, then $7^x> 16x^4$. So if $x>4$ and $y>3$, then $(7^x+3^y)> 16x^4+4y^2$, a contradiction. Case 1: $x\le 4$. If $x=1$, then $3^y+7\le 16+4y^2$. So $y<4$. Now $3^3+7=34\nmid (16+4\cdot 3^2)=52$. Thus, $y<3$. If $y=1$ or $y=2$, then $2(3^y+7)=16+4y^2$, which implies $7-3^y\mid 2$, so $y=2$. Indeed, $\boxed{(a,b)=(2,4)}$ is valid. If $x=2$, then $3^y+49\le 256+4y^2$. So $y<6$. If $y=5$, then LHS is $292$ and RHS is $356$, contradiction. If $y=4$, then LHS is $130$ and RHS is $320$, contradiction. Oops I forgot $y$ is even. Much less casework. If $y=2$, then we get $58\mid 272$, contradiction. If $x=3$, then $3^y+343\le 1296+4y^2$. So $y<7$. If $y=6$, then $1072\mid 1440$, contradiction. If $y=4$, then $424\mid 1360$, contradiction. If $y=2$, then $352\mid 1312$, contradiction. If $x=4$, then $2401+3^y\le 4096+4y^2$. So $y<7$. If $y=6$, then $3130\mid 4240$, contradiction. If $y=4$, then $2482\mid 4160$, contradiction. If $y=2$, then $2410\mid 4112$, contradiction. Case 2: $x>4$ and $y=2$. So we have $7^x+9\mid 16x^4+16$. So $x<5$, a contradiction. We have exhausted all cases, so we are done.

Note that $7^a-3^b$ is even, so $a^4+b^2$ is even. Thus, $a^4,b^2$ have same parity which means $a,b$ have same parity. $~$ If $a,b$ odd then $7^a-3^b$ is divisible by $4$ but $a^4,b^2\equiv 1\pmod 4$ so $a^4+b^2\equiv 2\pmod 4.$ $~$ Thus, $a,b$ are even. Let $a=2a_1,b=2b_1$ to get $49^{a_1}-9^{b_1}\equiv 0 \pmod 8$ which implies $16a_1^4+4b_1^2\equiv 0 \pmod 8$ so $b_1$ is even. Thus, let $b=4b_2.$ $~$ We have $49^{a_1}-81^{b_2}=(7^{a_1}-9^{b_2})(7^{a_1}+9^{b_2}).$ Note that the first factor is a difference of two odd but distinct integers, so it's at least two. Thus, $7^{a_1}+9^{b_2}\le 8(a_1^2+b_2^2)$ $~$ Easy bounding gives $9^k>8k^2$ and so $7^{a_1}\le 8a_1^2$ implying that $a_1=1,2$ or $3.$ Then, more bounding gives the solution $(1,1).$ Thus, $(a,b)=(2,4)$ is the only solution

. Clearly $a$ and $b$ have the same parity. Then, suppose that $a,b\equiv 1\pmod 2$. Then we have that $a^4+b^2\equiv 2\pmod 4$ while $7^a-3^b\equiv 0\pmod 4$, a contradiction. Then we may set $a=2x$ and $b=2y$ to obtain that \[7^{2x}-3^{2y}\mid 16x^4+4y^2\implies 7^x+3^y\mid 16x^4+4y^2\implies 7^x+3^y\le 16x^4+4y^2.\]Then by bounding we obtain that $x\le 4$ and $y\le 6$. Finally, plugging into our original equation gives that only $(a,b)=\boxed{(2,4)}$ works (we omit the calculations).

Obviously $a,b$ are the same parity. If they are both odd then $a^4+b^2 \equiv 2 \pmod 4$ and $7^a-3^b \equiv 0 \pmod 4$, a contradiction. Thus $a=2x$ and then $b=2y$. Thus $$7^{2x}-3^{2y}|16x^4+4y^2 \implies (7^x+3^y)(7^x-3^y)|16x^4+4y^2 \implies |(7^x-3^y)|(7^x+3^y) \le 16x^4+4y^2 \implies 7^x+3^y \le 8x^4+2y^2.$$Next note that $3^y \ge 2y^2+1$ for all $y \ge 1$, so we have that $7^x+1 \le 8x^4$. This inequality gives $x=1,2,3$. If $x=1$ then we have that $3^y \le 2y^2+1$ or $y=1,2$, and we get the solution $(x,y)=(1,2)$ so $(a,b)=(2,4)$. If $x=2$, then $3^y \le 79+2y^2$, so $y=1,2,3,4$ upon checking gives no solution. If $x=3$, then $3^y \le 305+2y^2$ which gives $y=1,2,3,4,5$ which upon checking gives no solutions, so the sole solution is $(a,b)=(2,4)$.

Missed the obvious bounding at the end :c (oops this looks like asdf's sol) The key claim is that Claim: $2|a$ and $4|b$. Proof: Since $7^a - 3^b$ is even, $a$ and $b$ must have the same parity. Thus $7^a - 3^b \equiv 0 \pmod{4}$ and so $2|a, b$. Then $7^a-3^b \equiv 1 - 1 = 0 \pmod{8}$ and so $4|b$. Let $a = 2c$ and $b = 4d$. We can rewrite the original equation as $$7^{2c}-3^{4d} \mid (2c)^4 + (4d)^2,$$or, $$(7^c+3^{2d})(7^c-3^{2d}) \mid 16(c^4+d^2) \implies (7^c+3^{2d}) \leq 16(c^4+d^2).$$Bound. This is left as an exercise to the reader. The answer is $(a, b) = \boxed{(2, 4)}$. $\square$

ISL Marabot solve Clearly, $7^a-3^b$ is even. So, parity of $a,b$ must be same. Therefore, $4|7^a-3^b$. Which gives, $a,b$ are even. Let, $a= 2x, b = 2y$. So, $(7^x+3^y)(7^x-3^y)|16x^4+4y^2$ Here, $7^x- 3^y$ is even and not $0$. So, $(7^x+3^y)|8x^4+2y^2$ $\implies 7^x+3^y \leq 8x^4+2y^2$ But, $3^y > 2y^2 \ \forall y\in \mathbb N$. So, $7^x > 8x^4$ which gives $x\leq 3$. And, for each $x$ we get $y < 3$. And by checking all of the pairs we can see that only $(x,y) = (1,2)$ works. Which gives the only solution $(a,b) = (2,4). \ \ \ \blacksquare$

We have that $7^a-3^b$ is even so we need the parities of $a,b$ to be the same. So we have that $4|7^a-3^b$ which implies that $a,b$ are even. We now let $a=2x$ and $b=2y$ so we have that $7^2x-3^2y=(7^x+3^y)(7^x-3^y)|16x^4+4y^2$ this implies that $7^x>8x^4$ else there is a contradiction so we have that $x \leq 3$ and $y<3$ so the only pair is $(x,y)=(1,2)$ which means that $(a,b)=(2,4)$ QED

Take mod $2$ $7^a-3^b \equiv 0 \pmod 2$ so $a^4+b^2 \equiv 0 \pmod 2$ $a+b \equiv 0 \pmod 2$ $a \equiv b \pmod 2$ Take mod $4$ $(-1)^a-(-1)^b \equiv 0 \pmod 4$ if $a$ and $b$ have the same parity, so $a^4+b^2 \equiv 0 \pmod 4$ and $a$ and $b$ are all even Let $a=2c, b=2d$ $49^c-9^d$ divides $16c^4+4d^2$ Take mod $8$ $49^c-9^d \equiv 0 \pmod 8$ So $16c^4+4d^2 \equiv 0 \pmod 8$ $4d^2 \equiv 0 \pmod 8$ $d^2 \equiv 0 \pmod 2$ $d \equiv 0 \pmod 2$ Let $d=2e$ $49^c-81^e$ divides $16c^4+16e^2$ $7^{2c}-9^{2e}$ divides $16c^4+16e^2$ $(7^c-9^e)(7^c+9^e)$ divides $16c^4+16e^2$ $|7^c-9^e|(7^c+9^e) \le 16c^4+16e^2$ $7^c+9^e \le 16c^4+16e^2$ For $e \ge 2$, $9^e > 16e^2$ because $e=2$ satisfies it and multiplying by $9$ is more than multiplying by $\frac{(e+1)^2}{e^2}$ so it will always satisfy For $c \ge 5$, $7^c > 16c^4$ because $c=5$ satisfies it and multiplying by $7$ is more than multiplying by $\frac{(c+1)^4}{c^4}$ so it will always satisfy So for $e \ge 2, c \ge 5$, $7^c+9^e > 16c^4+16e^2$ So to satisfy $7^c+9^e \le 16c^4+16e^2$, $e=1$ and $c=1,2,3,4$ Plug in all possible values of $c$ and $e$ to find that only $c=1, e=1$ works and results in $\boxed{a,b=(2,4)}$

The answer is $(a,b)=\boxed{(2,4)}$ only, which clearly works. Note that the LHS is a multiple of $4$, so the RHS must also be a multiple of $4$, so $a$ and $b$ are even. However, this makes the LHS a multiple of $8$, so $b$ is a multiple of $4$. Let $a=2m$ and $b=4n$. Then \[(7^m-9^n)(7^m+9^n)\mid 16(m^4+n^2).\]Since the first factor is even, we have \[7^m+9^n\mid 8(m^4+n^2)\rightarrow (7^m-8m^4)+(9^n-8n^2)\le 0.\] Let $f(m)=7^m-8m^4$ and $g(n)=9^n-8n^2$. Then we can see that $f(1)=-1$, $f(2)=-79$, $f(3)=-305$, $f(4)=353$, and $f(n)>353$ for all positive integers $n\ge 5$. We can also see that $g(1)=1$, $g(2)=49$, $g(3)=657$, and $g(n)>657$ for all positive integers $n\ge 4$. Therefore, we have to check $(m,n)=(1,1),(2,1),(2,2),(3,1),(3,2)$. If $(m,n)=(2,1)$, then $58\mid 8(17)$ which is not true. If $(m,n)=(2,2)$, then $130\mid 8(20)$ which is not true. If $(m,n)=(3,1)$, then $352\mid 8(82)$ which is not true. If $(m,n)=(3,2)$, then $424\mid 8(85)$ which is not true. If $(m,n)=(1,1)$, then $(a,b)=(2,4)$. QED.

This one's a few minute headsolve. Note that the LHS is divisible by 2, so a,b have same parity; in particular, if a,b odd then LHS is 0 mod 4, while RHS is 2 mod 4, contradiction, so we deduce a=2c,b=2d. Now the problem is evident because $7^c+3^d\mid 7^{2c}-3^{2d}\mid 16c^4+4d^2$, which it's obvious LHS increases faster so by obvious bounding we can manually check solutions that's only $2,4$.

As $7^a-3^b$ is even, $a, b$ are of the same parity; on the other hand, this implies $8 \mid 7^a - 3^b$, which forces $2 \mid a$ and $4 \mid b$ for mod $8$ reasons. On the other hand we must now have $7^{a/2} + 3^{b/2} \leq a^4+b^2$. By comparing the $a$ and $b$-terms, it suffices to check $(a, b) = (2, 4), (4, 4), (6, 4), (8, 4)$, of which only $(2, 4)$ works.

2007 N1 First of all, $a,b$ are of same parity implying that $ 8|7^a-3^b,$ so $2| a$ and $4|b$ , We can after that easily get that only $(2,4)$ works.

Quite a nice problem It's easy to find by $mod 4$ that $2 \mid a$ and $4 \mid b$ So suppose $a=2x$ and $b=4y$, we get that: $(7^x+9^y)(7^x-9^y) \mid 16(x^4+y^2)$ which means that $16(x^4+y^2) \ge 7^x+9^y$ It's obvious that $x \ge 5$ doesn't work and by casework($x=4, 3, 2, 1$) we get that the only solution is x=y=1 which means that $(a;b)=(2;4)$

bad formatting because copied from my mathdash solution note that 7^a - 3^b divides to, so a and b have the same parity, both odd causes 7^a - 3^b to be divsiible by 4 whereas a^4 + b^2 = 1 + 1 = 2 mod 4, so fail. then a,b, are both even, taking mod 8 gives a^4 + b^2 divides 8, so 4 | b. now factor (7^a - 3^b) = (7^x- 3^y)/2 *2(7^x + 3^y). now we then have 2(7^x + 3^y) <= 16x^4 + 4y^2, so 7^x + 3^y <= 8x^4 + 2y^2. now we must have one of the inequalities of x < 4, y < 2, true, otherwise we can just sum 7^x >8x^4, 3^y > 2y^2 and win. since y is even, we must have x < 4. x = 3 gives 343 + 3^y | 648 + 2y^2, clearly the left side is more than half the right side so we must have 3^y = 648 - 343 + 2y^2. this cleraly fails by size for y > 5, and y < 5 LHS is 2 digits RHS is 3. then y = 5 fails by direct computation. now we try x = 2, this gives 49 + 3^y <= 128 + 2y^2, for y > 5 we can win by size, y = 5 doesnt work by 2| y, y = 4 fails by direct computation, y = 3 fails by 2 | y, y = 2 fails by szie. now we try x = 1, or 7 + 3^y <= 8 + 2y^2, y > 2 fails by size, y = 2 works and yields the only solution that works, (2,4). it is trivial to see that this works.

By mod $4$ both $a, b$ are even. Then mod $8$ yields $4|b.$ Thus let $a=2x, b=4y.$ Then this yields $$(7^x-9^y)(7^x+9^y) \, | \, 16(x^4+y^2) \implies 16(x^4+y^2) \geq 7^x+9^y.$$Note that the RHS increases faster than the LHS for $x \geq 5, y \geq 2.$ If $y \geq 2,$ testing $x=1, 2, 3, 4$ yields no solutions. Therefore $y=1 \implies x = 1, 2, 3, 4.$ Only $x=1$ works meaning that the only solution is $(a, b) = (2, 4).$

We claim $(2, 4)$ is the only solution. Consider taking both sides mod $4$. We get that clearly the LHS is even, so $a$ and $b$ have equal parity. However, if they are both odd, the LHS is divisible by $4$ but the RHS is not. We perform the substitution $a=2x$, $b=2y$. Then we get $$(7^x-3^y)(7^x+3^y)|16x^4+4b^2.$$It must follow that since $7^x-3^y\neq 0$ and $7^x-3^x$ is even, it follows that $$2(7^x+3^y)|16x^4+4b^2.$$Now note that by this rule $x\leq 3$. Testing and bounding $y$ gives the desired result.