x^y+y^x=a
This is a general case. This question looks pretty easy in first look but when you try to solve it, it pretends like a hard question. First I will show you an example and then trick for general case.
Let we solve this example x^y+y^x=1649
1. This equation shows similarity with respect to x and y
⇒ Given curve is symmetric about x=y.
If (x,y) = (a,b) then (x,y) = (a,b) this must be true for all values of (x,y).
2. For this type of equation first we have to solve for equality (x=y)
⇒ x^x+x^x=1649
⇒ 2 x^x = 1649
⇒ x^x = 824.5
This is not possible since x must be an integer but this equation tells that x must be less than 5 (x<5).
From that we can say that any one value among (x,y) must be less than 5
3. Now we have to check for x=[1, 2, 3, 4]
In this case x,y both can not be even or both cannot be odd at a same time since 1649 is a odd number
4. Now let calculate for x=[1, 2, 3, 4] Manually
let x=1
⇒ 1^y+y^1=1649
⇒ y=1648
let x=2
⇒ 2^y+y^2=1649 (y<11 since 2^11>1649)
At y=10, 2^10+10^2=1124
Value of y must lie between (10,11)
y can not be an Integer for this equality
∴ x≠2
let x=3
⇒ 3^y+y^3=1649 (y<7 since 3^7>1649)
At y=6, 3^6+6^3=945
Value of y must lie between (6,7)
y can not be an Integer for this equality
∴ x≠3
let x=4
⇒ 4^y+y^4=1649 (y<6 since 4^6>1649)
At y=5, 4^5+5^4=1649
We got a perfect equality.
∴ x=4 is satisfying value when y=5
∴ (x,y)=(4,5)
5. From condition number 2 we can get remaining values of (x,y)
(x,y) have 4 sets of values
(1,1648), (4,5), (5,4), (1648,1)
These 4 are satisfying values for given equation.
6. Graph of given curve
For general case of x^y+y^x=a
For this you have simply have to follow simple steps as given below.
1. Try to solve at x=y
2. From this you will get an approximate or exact value of x or y.
3. By using error and trial method try some values but one value must be less than calculated value at x=y.
4. When you get values of x and y then replace that values and that will be another solution of this curve. For example you got (x,y)=(m,n) then (x,y)=(n,m) is another solution..
Tip: If a is even then x and y both are even or both are odd and if a is odd then one among x and y is odd and other must be even.
Lets us try some different example
x^y+y^x=5392
In this, both x and y are even or both are odd
1. This equation shows similarity with respect to x and y
⇒ Given curve is symmetric about x=y.
If (x,y) = (a,b) then (x,y) = (a,b) this must be true for all values of (x,y).
2. For this type of equation first we have to solve for equality (x=y)
⇒ x^x+x^x=5392
⇒ 2 x^x = 5392
⇒ x^x = 2696
This is not possible since x must be an integer but this equation tells that x must be less than 6 (x<6).
From that we can say that any one value among (x,y) must be less than 6
3. Now we have to check for x=[1, 2, 3, 4, 5]
In this case x,y both must be even or both must be odd at a same time since 5392 is an even number
4. Now let calculate for x=[1, 2, 3, 4] Manually
let x=1
⇒ 1^y+y^1=5392
⇒ y=5391
let x=2
⇒ 2^y+y^2=5392 (y<13 since 2^13>5392)
At y=12, 2^12+12^2=4240
Value of y must lie between (12,13)
y can not be an Integer for this equality
∴ x≠2
let x=3
⇒ 3^y+y^3=5392 (y<8 since 3^8>5392)
At y=7, 3^7+7^3=2530
Value of y must lie between (7,8)
y can not be an Integer for this equality
∴ x≠3
let x=4
⇒ 4^y+y^4=5392 (y<7 since 3^7>5392)
At y=6, 4^6+6^4=5392
We got a perfect equality.
∴ x=4 is satisfying value when y=6
∴ (x,y)=(4,6)
let x=5
⇒ 3^y+y^3=5392 (y<6 since 3^6>5392)
At y=5, 5^5+5^5=6250
Value of y must lie between (4,5)
y can not be an Integer for this equality
∴ x≠5
5. From condition number 2 we can get remaining values of (x,y)
(x,y) have 4 sets of values
(1,5391), (4,6), (6,4), (5391,1)
These 4 are satisfying values for given equation.
6. Graph of given curve
Lets us try one more example
x^y+y^x=512
In this, both x and y are even or both are odd
1. This equation shows similarity with respect to x and y
⇒ Given curve is symmetric about x=y.
If (x,y) = (a,b) then (x,y) = (a,b) this must be true for all values of (x,y).
2. For this type of equation first we have to solve for equality (x=y)
⇒ x^x+x^x=512
⇒ 2 x^x = 512
⇒ x^x = 256
This is not possible since 4^4=256. We got one set of (x,y) where both are 4, (x,y)=(4,4)
Now we have to try for other values where x is less than 4 (x<4)
3. Now we have to check for x=[1, 2, 3]
In this case x,y both must be even or both must be odd at a same time since 512 is an even number
4. Now let calculate for x=[1, 2, 3] Manually
let x=1
⇒ 1^y+y^1=512
⇒ y=511
let x=2
⇒ 2^y+y^2=512 (y<9 since 2^9=512)
At y=8, 2^8+8^2=320
Value of y must lie between (8,9)
y can not be an Integer for this equality
∴ x≠2
let x=3
⇒ 3^y+y^3=512 (y<6 since 3^6>512)
At y=5, 3^5+5^3=368
Value of y must lie between (5,6)
y can not be an Integer for this equality
∴ x≠3
5. From condition number 2 we can get remaining values of (x,y)
(x,y) have 4 sets of values
(1,511), (4,4), (511,1)
These 3 are satisfying values for given equation.
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