Solve a! + 5^b = 7^c for integers

[MyeonggyuRyu]

if a>=5 then a! and 5^b is a multiple of 5 (=0 (mod 5))therefore LHS is a multiple of 5. But RHS=7^c cannot be a multiple of 5 unless c=0 which cannot satisfy the equation
therefore no solution exists.

[SureshChoudhary-vhhq]

As a 10th grade student i dont know what mod is so i just worked it out my own and surprisingly I got the answer.
a! =7^c-5^b
The unit digit of 7^c-5^b cannot be zero
And from 5! Onwards every factorial has 0 in its unit place and 2, 3, 4 factorials are even numbers
So 7^c-5^b should be even
i.e. we need 5^b as odd
Therefore b=0, 1, 2 no need of further numbers since we need c>=b
a! =7^1-5^0=6, a=3
a! =7^1-5^1=2, a=2
a! =7^2-5^2=24, a=4
Therefore,
(a, b, c) =(2, 1, 1) (4, 2, 2) (3, 0, 1)

Idk about this plzz let me know if I'm wrong

[grotiuscollegedelftnlphysi]

What device or software do you use to write your formulas on-screen?

[jwy]

for the first part, couldnt you just say that if a is greater or equal to five the left side is divisible by five, but a power of 7 isnt, therefore a<5?

[pedrojose]

I dd it with boundaries and contradictions.
It is easy to see that a>=2 as 7^c is odd and c>=1 as a! + 5^b>=2.
b=0
If a>=7 then a!=7k, k integer and 7K+1 = 7^c so 7 | 7k+1, contradiction.
For 1< a < 7 it is easy to check that the only solution is (3, 0, 1).
b>=1
If a>=5 a!=5w, w integer so 5(w+5^(b-1)=7^c then 5 | 7^c, contradiction
So 1< a < 5
a= 2 it is eay to see that (2, 1, 1) is a solution. It is unique.
2 + 5^b =7^c, b, c>1 ==> 7^c=7 mod10 then c = 1 +4s, s integer
as 5^b=25 mod100 for b>1 then 7^c =27 mod 100.
But ord 7 mod100=4 as 7^4=49*49=(50-1)*(50-1)=2401
so (7^4k+1)=07 mod 100 and 2 + 5^b<>7^c for b>1.
then (2, 1, 1) is unique for a=2.
for a=3 we have found that (3, 0, 1) is a solution. And it is unique for a=3 as we will show.
for b=1 6 + 5= 7^c, no way.
For b>=1 7^c= 31 mod100 and 7^c=1 mod 10.
So c= 4m, m integer and m>0 as c>0.
So 7^c=1 mod100<>31 mod 100.
for a=4 24 + 5^b = 7^c so (4, 2, 2) is a solution and it is unique as we will show.
7^c=9 mod 10. So c= 2 + 4n, n integer. so c is even
If b is even, b=2u and c=2v, u,v integers.
(7^v+5^u)(7^v-7û)=24 as 5^u + 7^v>=12 the only solution is
7^v+5^u=12 and 7^v-5^u=2 and b=2 and c=2 as we have already found
If c is odd and c>1
7^c= 149 mod 1000 But as 7^c= 49 mod100 c= 2+4m, m integer.
7^c can be = 49, 649, 249, 849 or 449 mod1000 <> 149 mod 1000
So (3, 0, 1); (2, 1, 1) and (4, 2, 2) are the only solutions.

[unknown_]

Hi. Can you prompt how to plot the function {x}={y} on graph?

[mcwulf]

Quicker to start taking mod 5 then we have a! == 7^c (mod 5) which is impossible for a >= 5 as a! == 0 (mod 5). So a € {1, 2, 3, 4}.

But that's the easy bit. The hard part is to prove as you did, that the basic solution for each value of a is the only solution 👍

[yoav]

Short solution: if a>=5 there are no sol, because 5 can not divide 7^c. If a=0, 1 we have 5^b=7^c-1 but 6 divides 7^c-1 so no solutions. If a=2 for b=0 no sol for b=1c=1 is solution, for b>=2 we have 7^c=2(mod25) which can not be because 7^c=7, 24, 18, 1(mod25). For a=3 b=0 c=1 is solution, if b=1 no sol and for b>=2 we get 7^c=6(mod25) and no solutions. For a=4 we have 24+5^b=7^c. If we take mod5 we see c is even(4n+2)=2k and if we take mod7 we see b is even(6n+2)=2m. So we get: 24=(7^k-5^m)(7^k+5^m) so we have 2 cases: 7^k-5^m=4 7^k+5^m=6 no sol or 7^k-5^m=2 and 7^k+5^m=12 sòl k=1 and m=1 soa=4 b=c=2.( if b=0 we have a!+1=7^c which only works for a=3 c=1 as found)

[АзиретАкматбеков-йм]

Dear letsthinkcritically
I want to propose one question which looks interesting but I don't know how to do it. Could you send me the details?

[swenji]

This is kind of absurd to consider the powers of 7 mod 10 when a≥5 rather than just saying that a! cannot be a multiple of 5 and therefore a<5

[晓阳-dp]

for my opinion，
1）a！for a≧7，a！and 7^c ≡0（mod7）but 5^b not ≡0（mod7），so a cannot larger or equal to 7，
2）a！=1，2，6，24，120，...，last digit only 1，2，4，6，5^b last digit be 5 or 1，7^c last digit be 7，9，3，1

[arlethhernandez]

What if is a!+b!=c! What will happend in that case?

[محمدنیماکاظمی]

Here is my advise for you : If you want to attract more people and increase the number of subscribers you should solve harder problems than this in your channel.
By the way... thanks for your nice videos.