New Blogpost, after decades:-

INMO completed finally!!! I will write my experience/all other we do in the blog ( if I qualify/get merit awardee ) i.e making an excuse of not writing.

During this period, I and Anand ft Arpan Bhaiya did a lot of G-Solves, like seriously a lot. ( Fun fact:- Anand didn't prepare anything this year and he solved 3 problems and some partials. So he is too pr0. ).

Anyways, here's basically a collection of some modular bash problems, which one should try to motivate himself, they are RMO level. These problems are trivialised by zsig i.e why RMO level.

If you want more INMO level problems do check out https://artofproblemsolving.com/community/c1953209_olypd. I created it just to hide my frivolous forum posts (:P) but it's growing, and free of spam!

Some Problems:- ( If you want the source, feel free to ask in comments I am just to lazy to type the source), one can take this as an High  RMO level compilation mock(?). If you want to attempt this as a mock, then 3hrs is perfect!

1. Determine all pairs ($m,n$) of natural numbers for which $4^m + 5^n$ is a perfect square.

2. Find the sum of the numbers written with two digits $\overline{ab}$ for which the equation $3^{x + y} =3^x + 3^y + \overline{ab}$ has at least one solution $(x, y)$ in natural numbers.

3. Determine all pairs of integers $(x, y)$ that satisfy equation $(y - 2) x^2 + (y^2 - 6y + 8) x = y^2 - 5y + 62$.

4.Solve, in the positive integers, the equation $5^m + n^2 = 3^p$

5. Find all positive integers $x,y,z$ such that $7^x + 13^y = 8^z$

6. Find all integer solutions of the equation $14^x - 3^y = 2015.$

7.Let $\mathbb{N}$ denote the set of nonnegative integers. Find all solutions $(q, r, w) \in \mathbb{N}^3$ to the equation $$2013^q+2014^w=2015^r.$$

Problem 7 is from Arpan Bhaiya's mock( a2048's sacred mock)

Solution 1: Let $4^m+5^n=z^2 \implies 5^n=(z^2-2^m)(z^2+2^m).$ Let $d=(z^2-2^m,z^2+2^m) \implies d=1\implies z^2-2^m=1, z^2+2^m=5^n  \implies 5^n-1=2^{m+1}.$

Now using zsigmondy, we get that for $n>1,$ there will be a prime divisor dividing $5^n-1^n$ which is not $2.$

Hence $n=1 \implies m=1.$

So the only solution is $(m,n)=(1,1).$

Solution 2: Note that if both $x,y$ are $0$ or $1$ then it's not possible. Hence one of the variables is $>1$, wlog say $x.$ If $y$ is $0$ then not possible, since $3^x=3^x+1+...$ (not possible).So $y \ge 1$ and $x \ge 2.$

Now note that $3^5> \overline{ab}.$ So $3^x+3^y+3^5 >3^{x+y}.$ For $x,y \ge 3.$ The LHS will become too big and will start growing faster than RHS.

So only pair that can satisfy are $(x,y)=\{(2,3),(2,2),(2,1)\}$ ( we assumed $x\ge y$).

Verifying we get $(x,y)=(2,3)$ doesn't satisfy. For $(x,y)=(2,2),$ we get $\overline{ab}= 63$ and for $(x,y)=(2,1),$ we get $\overline{ab}= 15.$

By assuming $y\ge x,$ we get the same solution, so the numbers satisfying are $63,15.$

Hence answer is $\boxed{78}$

Solution 3: Expand it and we get

$$yx^2-2x^2+xy^2-6yx+8x-y^2+5y-62=0.$$ Now, we will make perfect squares and stuff,

we get $$[yx^2-2yx+y]+[4y-y^2-4]+[xy^2-4yx+4x]+[-2x^2+4x-2]=0$$ .

[This step looks simple, but it took a lot of time ]

Simplifying, we get $$y(x-1)^2-(y-2)^2+x(y-2)^2-2(x-1)^2=56 \implies (y-2)(x-1)(x+y-3)=56.$$ This is looking nice, we have to now find factors $a,b$ such that $a\cdot b\cdot (a+b)=56.$ For simplicity let's look at all the $3$ factor tuples such $xyz=56.$

Those are $(1,1,56);(1,2,28);(1,4,14);(1,7,8);(2,2,14);(2,4,7).$

Looking at these, the only 3 tuple which satisfy even if we introduce negative integers is $(1,7,8).$

Since negative integers are allowed , the three tuples which satisfy $a\cdot b (a+b)=56,$ where $y-2=a,x-1=b,x+y-3=56$ are

$$(1,7,8) \implies x=8,y=3$$ $$(7,1,8) \implies x=2,y=9$$ $$(7,-8,-1) \implies x=-7,y=9$$ $$(1,-8,-7) \implies x=-7,y=3$$ $$(-8,1,-7) \implies x=2,y=6$$ $$(-8,7,-1) \implies x=8,y=-6$$

So solutions are $\boxed{(x,y)=(8,3),(2,9),(-7,9),(-7,3),(2,6),(8,-6).}$ And we are done!

Solution 4: Clealy $n$ is even.

Now taking $\mod 4$ we get, $P$ even. Take $p=2k\implies 5^m=(3^k-n)(3^k+n)$.Letting $(3^q-n,3^q+n)=d \implies d=1\implies 5^m+1=2\cdot 3^k.$

By zigmondy, we get that for any $m>1,$ there will be another prime divisor, which divides $5^m+1$ and not $5+1=6.$

Hence $(m,n,p)=(1,2,2)$

Solution 5: Note that $z>0$. So taking $\mod 8,$ we get $x$ is odd and $y$ is even. If $y=0\implies$ by zsig, the only solution is $(x,z)=(1,1).$

If $y>0 \implies x \equiv 3 \mod 12, z \equiv 3 \mod 4.$ Letting $x=3k \implies 2^{3z}-7^{3k}=13^y \implies (2^z-7^k)(7^{2k}+2^z\cdot 7^k+2^{2z})=13^y.$

Letting $d=(2^z-7^k,7^{2k}+2^z\cdot 7^k+2^{2z})\implies d|2\cdot 2^z\cdot 7^k\implies 2^z-7^k =1.$ By zsig, the only sol is $(z,k)=(3,1) \implies y=2.$

So the solutions are $\boxed{(x,y,z)=(1,0,1),(3,2,3)}$

Solution 6: Taking $\mod 9,$ we get $x\equiv 3 \mod 6.$ Taking $\mod 13,$ we get $y\equiv 0 \mod 3.$

Then factorize and case bash.

Solution 7: $\boxed{(q, w, r) = (0, 1, 1)}$

Note that $r>0$ .Also, note that $w>0$, (if not then $2013^q$ is odd, $2014^w=1$ and $2015^r$ is odd, not possible as RHS is even and LHS is odd

Claim: $q=0$

Proof: let's say for contradiction , assume that $q /ge 1$ then $2013^q \equiv 0 \pmod 3$ and $2014^w \equiv 1 \pmod 3.$So we have $2013^q + 2014^w \equiv 1 mod 3 \implies r$ is even. Let $r=2b$ .Now, let us look at the unit digits,we see that $2013^q$ ends with $3,9,7,1$ and $2014^w$ ends with $4,6$ and $2015^r$ ends with $5$.

By calculations, we get that q must be even , hence let $q=2a$.

Hence we have $$2013^{2a} + 2014^w =2015^{2b} \implies 2014^w=2015^{2b}-2013^{2a}$$

$$\implies 2014^w= (2015^b-2103^a)(2015^b+2103^a) \implies 2^w \cdot 19^w \cdot 53^w = (2015^b-2103^a)(2015^b+2103^a)$$

Let $\alpha=(2015^b+2103^a)$ and $\beta=(2015^b-2103^a)$.

Clearly $\alpha>\beta$.let $d=\gcd(\alpha,\beta)$

Note that $d\mid 2014^w$ and $d\mid (\alpha +\beta)=2\cdot 2015^b \implies d=1$ or $2$ .

Now, since $\alpha>\beta$ , and $d= 1$ or $2$, we get that $v_{53}(\alpha)=w$ and $v_{19}(\beta)=w$.Now, in $\alpha , 2015^b \equiv 1 \pmod {53}$ , since $2015 \equiv 1 \pmod {53} .$

Since $2015^b - 2013^a \equiv 0 \pmod 53 \implies a$ is odd .

In $\beta, 2015^b \equiv 1 \pmod {19}$ , since $2015 \equiv 1 \pmod {19}.$ Now, since $a$ is odd $\implies 2013^a \equiv -1 \pmod {19} \implies \beta \equiv 2 \pmod {19}   \implies  v_{19}(\beta)=w=0$ , a contradiction, since in the first observation,we noted that $w>0$ .Hence $q=0$

Main Proof: So, we have $1+2014^w=2015^r$ . Note that $r$ is odd since $1 \equiv 1 \pmod 3, 2014\equiv 1 \pmod 3$. Hence $2014^w=2015^r-1=2014(2015^{r-1} +\dots +1).$ . But note $2015^{r-1} +\dots +1$ ( since $r$ is odd) .Hence $w=1$ .

Hence $2013^q+2014^w=2013^0+2014^1= 1+2014 =2015 \implies r=1$. And we are done!

Chalo that's it for this time :P. Do tell me how much time it took you all to solve!