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Question Number 212515 by efronzo1 last updated on 16/Oct/24

  The numbers of pairs of natural      numbers (x,y) with x,y ≤ 33 that      satisfy 5 ∣ 3^x^(y−1)   + 2^y^(2x)   is ...     (A) 295    (B) 296       (C) 297      (D) 298   (E) 299

Thenumbersofpairsofnaturalnumbers(x,y)withx,y33thatsatisfy53xy1+2y2xis...(A)295(B)296(C)297(D)298(E)299

Commented by A5T last updated on 16/Oct/24

I′m getting 313 pairs, do you have any solution  to this?

Imgetting313pairs,doyouhaveanysolutiontothis?

Answered by A5T last updated on 16/Oct/24

3^x^(y−1)  +2^y^(2x)  ≡(−2)^x^(y−1)  +2^y^(2x)  ≡0(mod 5)  Note that when x(=2k) is even and y≥3  (−2)^x^(y−1)  +2^y^(2x)  =((−2)^2^(y−1)  )^k^(y−1)  +2^y^(4k)  ≡0(mod 5)  ⇒2^y^(4k)  ≡4(mod 5)  ⇒y^(4k) −2≡0(mod 4)⇒y^(4k) ≡2(mod 4)     →←  So, x(=2k+1) is odd when y≥3  when y=1;3^x^(y−1)  +2^y^(2x)  ≡ 3+2≡0(mod 5) for any x  ⇒(x,y)=(1,1),(2,1),(3,1),...,(33,1)⇒33 solutions  when y=2; 3^x^(y−1)  +2^y^(2x)  ≡3^x +2^4^x  ≡3^x +1≡0(mod 5)  ⇒3^x ≡4(mod 5)⇒y=2∧x=4q+2⇒8 solutions  when y≥3, we know that x is odd  when x=1; 3^x^(y−1)  +2^y^(2x)  ≡3+2^y^2  ≡0(mod 5)  ⇒2^y^2  ≡2(mod 5)⇒y^2 ≡1(mod 4)⇒y=4l+1 or 4l+3≥3  ⇒(x,y)=(1,3),(1,5),...,(1,33)⇒16 solutions  Suppose y(=2m+1)≥3 is also odd and x≥3  3^x^(y−1)  +2^y^(2x)  =3^((2k+1)^(2m) ) +2^((2m+1)^(4k+2) )   ⇒2^((2m+1)^(4k+2) ) ≡2^((2k+1)^(2m) ) (mod 5)  ⇒(2m+1)^(4k+2) ≡(2k+1)^(2m) (mod 4)  ⇒1+(2m)^1 (4k+2)≡1+(2k)(2m) (mod 4)  which is always true for all m and k(≥1)  ⇒16×16=256 solutions  When y(=2m) is even.  ⇒3^((2k+1)^(2m−1) ) +2^((2m)^(4k+2) ) ≡0(mod 5)  ⇒(2m)^(4k+2) ≡(2k+1)^(2m−1) (mod 4)     →←  ⇒Total number of solutions=256+16+8+33  =313

3xy1+2y2x(2)xy1+2y2x0(mod5)Notethatwhenx(=2k)isevenandy3(2)xy1+2y2x=((2)2y1)ky1+2y4k0(mod5)2y4k4(mod5)y4k20(mod4)y4k2(mod4)→←So,x(=2k+1)isoddwheny3wheny=1;3xy1+2y2x3+20(mod5)foranyx(x,y)=(1,1),(2,1),(3,1),...,(33,1)33solutionswheny=2;3xy1+2y2x3x+24x3x+10(mod5)3x4(mod5)y=2x=4q+28solutionswheny3,weknowthatxisoddwhenx=1;3xy1+2y2x3+2y20(mod5)2y22(mod5)y21(mod4)y=4l+1or4l+33(x,y)=(1,3),(1,5),...,(1,33)16solutionsSupposey(=2m+1)3isalsooddandx33xy1+2y2x=3(2k+1)2m+2(2m+1)4k+22(2m+1)4k+22(2k+1)2m(mod5)(2m+1)4k+2(2k+1)2m(mod4)1+(2m)1(4k+2)1+(2k)(2m)(mod4)whichisalwaystrueforallmandk(1)16×16=256solutionsWheny(=2m)iseven.3(2k+1)2m1+2(2m)4k+20(mod5)(2m)4k+2(2k+1)2m1(mod4)→←Totalnumberofsolutions=256+16+8+33=313

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