Solution: Note $(2^{107+140k})^3 + (2^{80+105k})^4 + (2^{64+84k})^5 = (2^{46+60k})^7$ satisfies the condition for all $k\geq 0$.

how did u reach this ?

iamsidharthnair wrote: how did u reach this ? It was obvious that getting one solution was enough. Then $2^a + 2^a + 2^{a+1} = 2^{a+2}$ was the equality I noticed, then was simple.

$ (k^{140} * 3^{100+7k})^3 + (k^{105} * 3^{75})^4 + (k^{84} * 3^{60})^5 = (k^{60} * 3^{43+3k})^7 $ also satisfies for all $k \geq 0$

Actually I solved a question just like this earlier this week. I think it was in Canada 1991 paper.

How many marks will I get for just writing that we can obtain many solutions after we get one, but not writing that first one?

Yet another solution: Take $$a=3^{20k},b=3^{15k},c=3^{12k}$$where $k$ is a natural number. Then $$a^3+b^4+c^5=3^{60k}+3^{60k}+3^{60k}=3^{60k+1}$$So it suffices to show that there are infinitely many integers $k \in \mathbb{N}$ such that $$60k+1 \equiv 0 \pmod{7} \Leftrightarrow 4k \equiv -1 \equiv 20 \pmod{7} \Leftrightarrow k \equiv 5 \pmod{7}$$which is obviously true. Hence, done. $\blacksquare$ Motivation: The motivation is the fact that $3^n+3^n+3^n=3^{n+1}$. Now taking $n=\text{lcm}(3,4,5) \times k$ easily gives the result.

$a=x(x^3+y^4+z^5)^{100}\\b=y(x^3+y^4+z^5)^{75}\\ c=z(x^3+y^4+z^5)^{60}\\d=(x^3+y^4+z^5)^{43}$

I have an alternative approach Since a,b,c,d are natural numbers b=ka ( for some positive rational number' k') c=la ( for some positive rational number 'l') d=ma( for some positive rational number 'm') Now let us evaluate the given expression by putting the values of a,b,c and d We get a^3+(ka)^4+(la)^5=(ma)^7 [ We need to confirm if our assumption is correct or wrong] Since a is non zero, we can divide both sides by a^3. We get 1+k^4a+l^5a^2=m^7a^4 Now again divide by a^2 on both sides We get finally m^7a^2-l5-(k^4/a)-1/(a^2)=0 This is a quadratic in 'a' . Let us check the discriminant of it. Discriminant is 4m^7(l^5+k^4/a +1/a^2) Now m,k and a are all positive . So discriminant is positive Meaning the quadratic in 'a' produces real solutions of a Ie . There are infinitely many values of 'a' So there are infinitely many values of b,c and d because b=ka c=la d=ma Hence we have infinitely such tuples

TheDarkPrince wrote: Show that there are infinitely many tuples $(a,b,c,d)$ of natural numbers such that $a^3 + b^4 + c^5 = d^7$. A completely different solution. Let $k$ be an integer such that $3|k, 4|k, 5|k, 7|k+1$, By CRT, there exist infintely many such numbers. Let $a^3=3^k, b^4=3^k, c^5=3^k, d^7=3^{k+1}$. Then the equation stands.. $3^k+3^k+3^k=3^{k+1}$, which is true for any $k$. Hence, there are infinitely many such numbers....

For all $t \in \mathbb{Z}^{+}$, we have, $$(3^{140t+100})^3+(3^{105t+75})^4+(3^{84t+60})^5=(3^{60t+43})^7$$

Just a thought: Do there exist infinite solutions if we add the extra condition $gcd(a,b,c,d)=1$?

Mathotsav wrote: Just a thought: Do there exist infinite solutions if we add the extra condition $gcd(a,b,c,d)=1$? You mean pairwise co prime or all relatively prime to each other?

ShamimAkhtar212 wrote: TheDarkPrince wrote: Show that there are infinitely many tuples $(a,b,c,d)$ of natural numbers such that $a^3 + b^4 + c^5 = d^7$. A completely different solution. Let $k$ be an integer such that $3|k, 4|k, 5|k, 7|k+1$, By CRT, there exist infintely many such numbers. Let $a^3=3^k, b^4=3^k, c^5=3^k, d^7=3^{k+1}$. Then the equation stands.. $3^k+3^k+3^k=3^{k+1}$, which is true for any $k$. Hence, there are infinitely many such numbers.... Doesn't CRT guarantee an unique solution?..

^ It did one year ago, Now its a fairy tale...

TheDarkPrince wrote: Show that there are infinitely many tuples $(a,b,c,d)$ of natural numbers such that $a^3 + b^4 + c^5 = d^7$. The problem is spacial case.. general problem Show that there are infinitely many tuples $(x_1,x_2..,x_n)$ of natural numbers such that $x_1^t+x_2^{t+1}+...+x_{n-1}^{t+n-2}=x_n^m$.($ m,n,t$ natural numbers and fixed also $gcd(m,lcm(t,..,t+n-2))=1$) soloution Let $x_i=(n-1)^{k\frac{lcm(t,t+1,...,t+n-2)}{t+i-1}} $ then we have $x_n^m=x_1^t+x_2^{t+1}+...+x_{n-1}^{t+n-2}=(n-1)^{k.lcm(t,t+1,...,t+n-2)}.(n-1)=(n-1)^{k.lcm(t,t+1,...,t+n-2)+1}$ So it is enough to prove it there is$k$ such that $$k.lcm(t,t+1,...,t+n-2)+1\equiv 0 \pmod{m}$$but be according to the case, Bezout’s theorem is right!

AZOT1 wrote: TheDarkPrince wrote: Show that there are infinitely many tuples $(a,b,c,d)$ of natural numbers such that $a^3 + b^4 + c^5 = d^7$. The problem is spacial case.. general problem Show that there are infinitely many tuples $(x_1,x_2..,x_n)$ of natural numbers such that $x_1^t+x_2^{t+1}+...+x_{n-1}^{t+n-2}=x_n^m$.($ m,n,t$ natural numbers and fixed also $gcd(m,lcm(t,..,t+n-2))=1$ soloution Let $x_i=(n-1)^{k\frac{lcm(t,t+1,...,t+n-2)}{t+i-1}} $ then we have $x_n^m=x_1^t+x_2^{t+1}+...+x_{n-1}^{t+n-2}=(n-1)^{k.lcm(t,t+1,...,t+n-2)}.(n-1)=(n-1)^{k.lcm(t,t+1,...,t+n-2)+1}$ So it is enough to prove it there is$k$ such that $$k.lcm(t,t+1,...,t+n-2)+1\equiv 0 \pmod{m}$$but be according to the case, Bezout’s theorem is right! AZOT1 the problem is spacial case.. ]General problem Show that there are infinitely many tuples $(x_1,x_2..,x_n)$ of natural numbers such that $x_1^{a_1}+x_2^{a_2}+...+x_{n-1}^{a_{n-1}}=x_n^m$.($ m,n,t$ natural numbers and fixed also $gcd(m,lcm(a_1,..,a_{n-1}))=1$ soloution Let $x_i=(n-1)^{k\frac{lcm(a_1,..,a_{n-1})}{t+i-1}} $ then we have $x_n^m=x_1^t+x_2^{t+1}+...+x_{n-1}^{t+n-2}=(n-1)^{k.lcm(a_1,..,a_{n-1})}.(n-1)=(n-1)^{k.lcm(a_1,..,a_{n-1})+1}$ So it is enough to prove it there is$k$ such that $$k.lcm(a_1,..,a_{n-1})+1\equiv 0 \pmod{m}$$but be according to the case, Bezout’s theorem is right!

math_pi_rate wrote: Yet another solution: Take $$a=3^{20k},b=3^{15k},c=3^{12k}$$where $k$ is a natural number. Then $$a^3+b^4+c^5=3^{60k}+3^{60k}+3^{60k}=3^{60k+1}$$So it suffices to show that there are infinitely many integers $k \in \mathbb{N}$ such that $$60k+1 \equiv 0 \pmod{7} \Leftrightarrow 4k \equiv -1 \equiv 20 \pmod{7} \Leftrightarrow k \equiv 5 \pmod{7}$$which is obviously true. Hence, done. $\blacksquare$ Motivation: The motivation is the fact that $3^n+3^n+3^n=3^{n+1}$. Now taking $n=\text{lcm}(3,4,5) \times k$ easily gives the result. Why do you claim that $k\equiv \pmod{7}$? You just found out a condition that $k$ must have a remainder $5$ when divided by $7$. So, we can take suitable $k's$ for the equation to hold. As the number of such 'suitable' $k's$ is infinite, therefore, there are infinite possible tuples that satisfy the equation.. am I correct? If you did prove that $k$ is always $\equiv \pmod{7}$, then can you please explain how did you do that?

This is classic problem. I saw a beautiful parameterized for this. Given $$\sum_{i=1}^n a_i^{p_i}=d^p$$Where $\gcd(lcm(p_1,p_2,\cdots,p_n),p)=1$ Then we can take natural numbers $\{b_n\}$ and set $$a_i=b_i\left(\sum_{i=1}^n b_i^{p_i}\right)^{\frac{Mk}{p_i}}$$Where $M=lcm(p_1,p_2,\cdots,p_n)$ and $k$ an integer such that $p\mid Mk+1$.

Generalization Show that there are infinitely many tuples $(a_1,a_2..,a_n)$ ,$(b_1,b_2..,b_m)$of natural numbers such that $$\sum_{i=1}^n a_i^{p_i}=\sum_{j=1}^n b_j^{q_j}$$.($ p_i,q_j,n, m$natural numbers and fixed also $\gcd(lcm(p_1,p_2,\cdots,p_n),\gcd(lcm(q_1,q_2,\cdots,q_m))=1$ soloution Lemma: Show that there are infinitely many tuples $(x_1,x_2..,x_n)$ of natural numbers such that $x_1^{t_1}+x_2^{t_2}+...+x_{n-1}^{t-1}=x_n^m$.($ m,n,t$ natural numbers and fixed also $gcd(m,lcm(t_1,..,t_{n-1}))=1$ proof Then we can take natural numbers $\{b_n\}$ and set $$a_i=b_i\left(\sum_{i=1}^n b_i^{p_i}\right)^{\frac{Mk}{p_i}}$$Where $M=lcm(p_1,p_2,\cdots,p_n)$ and $k$ an integer such that $m\mid Mk+1$(according to the case, Bezout’s theorem )$\blacksquare$ Back to the main problem: $Case1: \min(m,n)=1$ equal to lemma. $Case1: \min(m,n)>1$ then according lemma This equations have infinitely many soloution in natural numbers: $$1)\sum_{i=1}^{n-1} a_i^{p_i}=\sum_{j=1}^1 b_j^{q_j}$$$$2)\sum_{i=n}^{n} a_i^{p_i}=\sum_{j=2}^m b_j^{q_j}$$ then according $(1),(2)$ we conclude that the equation$$\sum_{i=1}^n a_i^{p_i}=\sum_{j=1}^m b_j^{q_j}$$have infinitely many soloution in natural numbers. $\blacksquare$

Well, I used powers of $3$ as $3^{20(7k+5)},3^{15(7k+5)},3^{12(7k+5)}$ respectively as $a,b,c$ Also, as we see that $20(7k+5)+1$ is divisible by $7$, we will surely have some natural no. $t$ such that $d=3^t$

Used crt and darklords idea and got [2^(648+840k)]^3+[2^(486+630k)]+[2^(389+504k)]=[2^(278+360k)] for all non negative k

New solution. Consider the following construction. \[ (a,b,c,d) = (3^{100}\cdot n^{140}, 3^{75}\cdot n^{105}, 3^{60}\cdot n^{84}, 3^{43}\cdot n^{60}) .\]Clearly this works for any natural $n$ and we are done.